Circuits and Networks
Analysis and Synthesis
Fifth Edition
About the Authors
A Sudhakar is presently working as Principal at RVR & JC College of Engineering,
Guntur, Andhra Pradesh. He obtained his BTech (1978–83) from V R Siddartha
Engineering College affiliated to Acharya Nagarjuna University, and ME from Jadavpur
University, Kolkata (1983–85). Thereafter, he obtained his PhD from Andhra University,
Visakhapatnam, in 2002. He has 29 years of teaching and research experience in the
field of Electronics and Communication Engineering and has also worked as Associate
Professor at AIMST, Malaysia, during 2003–2005. Presently, he is a member of the Planning and Monitoring
Board and a Senate Member of Acharya Nagarjuna University. He is a member of various professional
societies like IEEE, SEMCEI, ISTE, and Fellow in IETE.
Dr Sudhakar has visited several countries like Malaysia, Singapore, Russia, and Sweden to present papers
in international conferences. He has published more than 50 papers in various national and international
journals and has co-authored several books, all published by McGraw Hill Education (India). He has
completed several projects and is working on various new research and sponsored projects funded by AICTE
and UGC, Govt of India. His fields of interest include Circuits, Signals and Systems, and Array Antennas.
He received the Prathibha Puraskar from T R Mahesh Charitable Trust, Guntur, for his services in the field
of Higher Education, Andhra Pradesh on 19th July, 2014. He has received a Certificate of Excellence for
his outstanding services, achievements, and contributions to the field of Electronics and Communication
Engineering by the IETE Vijayawada centre on the occasion of “IETE Foundation Day” on 2 November,
2014. Dr A Sudhakar, as Head of the Institution, received the Prestigious Indira Gandhi National Service
Scheme Award 2013–14 from the President of India for the NSS Unit of RVR & JC College of Engineering.
Shyammohan S Palli is presently Professor and Head, Department of Electrical and
Electronics Engineering, Sir C R Reddy College of Engineering, Eluru, affiliated to
Andhra University. Prior to his current assignment, he worked in different capacities,
first as Lecturer at Siddaganga Institute of Technology, Tumkur, in the year 1981 and
then, he was Lecturer at K L College of Engineering during 1981–87. From 1987–2001,
he worked at RVR & JC College of Engineering, Guntur, first as Assistant Professor and
then as Professor and Head.
He obtained his BTech in Electrical Engineering from the College of Engineering, Kakinada (JNTU), in
1978. He completed his MTech, specializing in Power Systems in 1980 from the same college. Since then
he has been associated with teaching and research and has over 34 years of experience. He has published
numerous papers in national and international journals and has authored several books on Networks and
Circuits which have found national acclaim. He has also received various sponsored projects funded by
AICTE. Presently, he is pursuing PhD in the area of Power System Control at Andhra University. His areas
of interest include Control Systems, Power Systems, Operation and Control.
Both the authors also have, to their credit, books specific for regional technical universities, such as for
Jawaharlal Nehru Technological University, published by McGrawHill Education (India).
Circuits and Networks
Analysis and Synthesis
Fifth Edition
A Sudhakar
Principal and Professor
RVR & JC College of Engineering
Guntur, Andhra Pradesh
Shyammohan S Palli
Professor and Head
Department of Electrical and Electronics Engineering
Sir C R Reddy College of Engineering
Eluru, Andhra Pradesh
McGraw Hill Education (India) Private Limited
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Circuits and Networks: Analysis and Synthesis, 5e
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Contents
Preface
xix
Chapter 1 Circuit Elements and Kirchhoff’s Laws
Learning Objectives
1.1 Voltage
1.2 Current
1.3 Power and Energy
Practice Problems linked to LO 2
Frequently Asked Questions linked to LO 2
1.4 The Circuit
Frequently Asked Questions linked to LO 3
1.5 Resistance Parameter
Practice Problems linked to LO 4
Frequently Asked Questions linked to LO 4
1.6 Inductance Parameter
1.7 Capacitance Parameter
Practice Problems linked to LO 6
Frequently Asked Questions linked to LO 6
1.8 Energy Sources
Frequently Asked Questions linked to LO 7
1.9 Kirchhoff’s Voltage Law
1.10 Voltage Division
1.11 Power in a Series Circuit
Practice Problems linked to LO 8
Frequently Asked Questions linked to LO 8
1.12 Kirchhoff’s Current Law
1.13 Parallel Resistance
1.14 Current Division
1.15 Power in a Parallel Circuit
Practice Problems linked to LO 9
Frequently Asked Questions linked to LO 9
Additional Solved Problems
PSpice Problems
Answers to Practice Problems
Objective-Type Questions
1
1
1
2
2
3
4
4
5
5
6
6
7
8
9
10
10
11
11
14
15
16
17
17
20
21
22
22
23
25
40
44
45
Chapter 2 Methods of Analysing Circuits
Learning Objectives
2.1 Introduction
2.2 Tree and Co-tree
2.3 Twigs and Links
Frequently Asked Questions linked to LO1
47
47
47
48
50
50
viii
Contents
2.4
2.5
2.6
Incidence Matrix (A)
Properties of Incidence Matrix A
Incidence Matrix and KCL
Frequently Asked Questions linked to LO2
Link Currents: Tie-Set Matrix
Frequently Asked Questions linked to LO3
Cut-Set and Tree Branch Voltages
Frequently Asked Questions linked to LO4
Mesh Analysis
Mesh Equations by Inspection Method
Supermesh Analysis
Practice Problems linked to LO5
Frequently Asked Questions linked to LO5
Nodal Analysis
Nodal Equations by Inspection Method
Supernode Analysis
Practice Problems linked to LO 6
Frequently Asked Questions linked to LO 6
Source Transformation Technique
Frequently Asked Questions linked to LO 7
Additional Solved Problems
PSpice Problems
Answers to Practice Problems
Objective-Type Questions
50
51
53
54
54
58
60
65
65
68
70
72
75
78
81
83
84
86
88
90
91
106
108
108
Chapter 3 Useful Theorems in Circuit Analysis
Learning Objectives
3.1 Star-Delta Transformation
Practice Problems linked to LO 1
Frequently Asked Questions linked to LO 1
3.2 Superposition Theorem
Practice Problems linked to LO 2
Frequently Asked Questions linked to LO 2
3.3 Thevenin’s Theorem
Practice Problems linked to LO 3
Frequently Asked Questions linked to LO 3
3.4 Norton’s Theorem
Practice Problems linked to LO 4
Frequently Asked Questions linked to LO 4
3.5 Reciprocity Theorem
Frequently Asked Questions linked to LO 5
3.6 Compensation Theorem
Frequently Asked Questions linked to LO 6
3.7 Maximum Power Transfer theorem
Practice Problems linked to LO 7
Frequently Asked Questions linked to LO 7
110
110
110
114
114
115
118
119
120
122
123
124
126
126
127
129
129
130
130
131
132
2.7
2.8
2.9
2.10
2.11
2.12
2.13
2.14
2.15
Contents
ix
Duals and Duality
Frequently Asked Questions linked to LO 8
3.9 Tellegen’s Theorem
Frequently Asked Questions linked to LO 9
3.10 Millman’s Theorem
Frequently Asked Questions linked to LO 10
Additional Solved Problems
PSpice Problems
Answers to Practice Problems
Objective-Type Questions
133
135
135
136
136
138
138
160
163
163
3.8
Chapter 4 Introduction to Alternating Currents and Voltages
Learning Objective
4.1 The Sine Wave
Practice Problems linked to LO 1
4.2 Angular Relation of a Sine Wave
4.3 The Sine Wave Equation
Practice Problems linked to LO 2
4.4 Voltage and Current Values of a Sine Wave
Practice Problems linked to LO 3
Frequently Asked Questions linked to LO 3
4.5 Phase Relation in a Pure Resistor
4.6 Phase Relation in a Pure Inductor
4.7 Phase Relation in a Pure Capacitor
Practice Problems linked to LO 4
Additional Solved Problems
Answers to Practice Problems
Objective-Type Questions
165
165
165
167
167
169
170
170
173
174
174
175
176
177
177
183
183
Chapter 5 Complex Impedance
Learning Objectives
5.1 Impedance Diagram
5.2 Phasor Diagram
Practice Problems linked to LO 2
5.3 Series Circuits
Practice Problems linked to LO 3
Frequently Asked Questions linked to LO 3
5.4 Parallel Circuits
Practice Problems linked to LO 4
5.5 Compound Circuits
Practice Problems linked to LO 5
Frequently Asked Questions linked to LO 5
Additional Solved Problems
PSpice Problems
Answers to Practice Problems
Objective-Type Questions
185
185
185
187
188
189
193
195
195
197
197
199
201
201
216
218
219
x
Contents
Chapter 6 Power and Power Factor
Learning Objectives
6.1 Instantaneous Power
Practice Problems linked to LO 1
6.2 Average Power
Practice Problems linked to LO 2
6.3 Apparent Power and Power Factor
Practice Problems linked to LO 3
Frequently Asked Questions linked to LO 3
6.4 Reactive Power
6.5 The Power Triangle
Practice Problems linked to LO 5
Additional Solved Problems
PSpice Problems
Answers to Practice Problems
Objective-Type Questions
221
221
221
222
223
224
225
226
228
228
229
229
230
240
242
242
Chapter 7 Steady-State AC Analysis
Learning Objectives
7.1 Mesh Analysis
7.2 Mesh Equations by Inspection
Practice Problems linked to LO 1
Frequently Asked Questions linked to LO 1
7.3 Nodal Analysis
7.4 Nodal Equations by Inspection
Practice Problems linked to LO 2
Frequently Asked Questions linked to LO 2
7.5 Superposition Theorem
Practice Problems linked to LO 3
Frequently Asked Questions linked to LO 3
7.6 Thévenin’s Theorem
Practice Problems linked to LO 4
Frequently Asked Questions linked to LO 4
7.7 Norton’s Theorem
Practice Problems linked to LO 5
Frequently Asked Questions linked to LO 5
7.8 Maximum Power Transfer Theorem
Practice Problems linked to LO 6
Frequently Asked Questions linked to LO 6
Additional Solved Problems
Frequently Asked Questions
PSpice Problems
Answers to Practice Problems
Objective-Type Questions
244
244
244
246
248
248
249
252
254
255
256
257
258
258
260
260
261
262
262
262
264
265
265
280
280
282
282
Contents
xi
Chapter 8 Resonance
Learning Objectives
8.1 Series Resonance
Frequently Asked Questions linked to LO 1
8.2 Impedance and Phase Angle of A Series Resonant Circuit
Frequently Asked Questions linked to LO 2
8.3 Voltages and Currents in a Series Resonant Circuit
8.4 Bandwidth of An RLC Circuit
8.5 The Quality Factor (Q) and its Effect on Bandwidth
8.6 Magnification in Resonance
Practice Problems linked to LO 3
Frequently Asked Questions linked to LO 3
8.7 Parallel Resonance
8.8 Resonant Frequency for a Tank Circuit
Frequently Asked Questions linked to LO 4
8.9 Variation of Impedance With Frequency
8.10 Q-Factor of Parallel Resonance
8.11 Magnification
8.12 Reactance Curves in Parallel Resonance
Practice Problems linked to LO 5
Frequently Asked Questions linked to LO 5
8.13 Locus Diagrams
Additional Solved Problems
PSpice Problems
Answers to Practice Problems
Objective-Type Questions
284
284
284
286
286
288
288
291
294
295
296
297
298
299
300
301
302
304
304
305
305
306
315
326
330
330
Chapter 9 Polyphase Circuits
Learning Objectives
9.1 Polyphase System
9.2 Advantages of a Three-Phase System
Frequently Asked Questions linked to LO1
9.3 Generation of Three-Phase Voltages
9.4 Phase Sequence
Frequently Asked Questions linked to LO3
9.5 Interconnection of Three-Phase Sources and Loads
Practice Problems linked to LO 4
Frequently Asked Questions linked to LO 4
9.6 Star-To-Delta and Delta-To-Star Transformation
Frequently Asked Questions linked to LO 5
9.7 Voltage, Current, and Power in a Star Connected System
Practice Problems linked to LO 6
Frequently Asked Questions linked to LO 6
9.8 Voltage, Current, and Power in a Delta-Connected System
Practice Problems linked to LO 7
Frequently Asked Questions linked to LO 7
331
331
331
332
332
332
334
335
335
339
340
340
344
344
348
348
349
353
353
xii
Contents
9.9 Three-Phase Balanced Circuits
9.10 Three-Phase-Unbalanced Circuits
Practice Problems linked to LO 8
Frequently Asked Questions linked to LO 8
9.11 Power Measurement in Three-Phase Circuits
Practice Problems linked to LO 9
Frequently Asked Questions linked to LO 9
9.12 Effects of Harmonics
9.13 Effects of Phase-Sequence
Practice Problems linked to LO 11
9.14 Power Factor of an Unbalanced System
Additional Solved Problems
PSpice Problems
Answers to Practice Problems
Objective-Type Questions
Chapter 10 Coupled Circuits
Learning Objectives
10.1 Introduction
10.2 Conductively Coupled Circuit and Mutual Impedance
10.3 Mutual Inductance
Practice Problems linked to LO 2
Frequently Asked Questions linked to LO 2
10.4 Dot Convention
Practice Problems linked to LO 3
Frequently Asked Questions linked to LO 3
10.5 Coefficient of Coupling
Frequently Asked Questions linked to LO 4
10.6 Ideal Transformer
Practice Problems linked to LO 5
Frequently Asked Questions linked to LO 5
10.7 Analysis of Multi-Winding Coupled Circuits
Practice Problems linked to LO 6
10.8 Series Connection of Coupled Inductors
Practice Problems linked to LO 7
Frequently Asked Questions linked to LO 7
10.9 Parallel Connection of Coupled Coils
Practice Problems linked to LO 8
Frequently Asked Questions linked to LO 8
10.10 Tuned Circuits
Practice Problems linked to LO 9
Frequently Asked Questions linked to LO 9
10.11 Analysis of Magnetic Circuits
Practice Problems linked to LO 10
Frequently Asked Questions linked to LO 10
10.12 Series Magnetic Circuit
354
358
368
369
369
379
379
379
383
384
384
385
408
410
411
413
413
413
414
414
416
416
417
419
419
419
421
422
427
427
428
430
430
431
431
432
433
433
434
438
439
440
441
441
441
Contents
10.13 Comparison of Electric and Magnetic Circuits
Practice Problems linked to LO 11
Frequently Asked Questions linked to LO 11
10.14 Magnetic Leakage and Fringing
10.15 Composite Series Circuit
10.16 Parallel Magnetic Circuit
Frequently Asked Questions linked to LO 13
Additional Solved Problems
Frequently Asked Questions
PSpice Problems
Answers to Practice Problems
Objective-Type Questions
xiii
442
444
444
444
445
446
446
446
462
462
464
464
Chapter 11 Transients
Learning Objectives
11.1 Steady State and Transient Response
Practice Problems linked to LO 1
Frequently Asked Questions linked to LO 1
11.2 DC Response of an R-L Circuit
11.3 DC Response of an R-C Circuit
11.4 DC Response of an R-L-C Circuit
Practice Problems linked to LO 2
Frequently Asked Questions linked to LO 2
11.5 Sinusoidal Response of an R-L Circuit
11.6 Sinusoidal Response of an R-C Circuit
11.7 Sinusoidal Response of an R-L-C Circuit
Practice Problems linked to LO 3
Frequently Asked Questions linked to LO 3
Additional Solved Problems
PSpice Problems
Answers to Practice Problems
Objective-Type Questions
466
466
466
467
467
467
470
472
475
477
481
484
487
492
494
494
507
510
510
Chapter 12 Fourier Method of Waveform Analysis
Learning Objectives
12.1 Introduction to Fourier Analysis
12.2 Trigonometric Series
12.3 Compact Trigonometric Fourier Series
Practice Problems linked to LO 1
Frequently Asked Questions linked to LO 1
12.4 Functions of any Period 2T
Practice Problems linked to LO 2
Frequently Asked Questions linked to LO 2
12.5 Complex Fourier Series
12.6 Amplitude and Phase Spectrums
12.7 The Frequency Spectrum
512
512
512
512
515
516
517
517
521
522
522
524
525
xiv
Contents
12.8
12.9
12.10
12.11
12.12
12.13
Practice Problems linked to LO 3
Frequently Asked Questions linked to LO 3
Fourier Transform
The Energy Spectrum
Frequently Asked Questions linked to LO 4
Fourier Transform of Power Signals
Fourier Transform of Periodic Signals
Practice Problems linked to LO 5
Properties of the Fourier Transform
Practice Problems linked to LO 6
Frequently Asked Questions linked to LO 6
Applications in circuit Analysis
Additional Solved Problems
Answers to Practice Problems
Objective-Type Questions
525
526
526
529
530
530
534
536
536
541
541
541
544
560
561
Chapter 13 Introduction to the Laplace Transform
Learning Objectives
13.1 Definition of the Laplace Transform
Frequently Asked Questions Linked to LO 1
13.2 Step Function
13.3 Impulse Function
13.4 Functional Transforms
13.5 Operational Transforms
Practice Problems linked to LO 2
Frequently Asked Questions linked to LO 2
13.6 Laplace Transform of Periodic Functions
Practice Problems linked to LO 3
Frequently Asked Questions linked to LO 3
13.7 Inverse Transforms
Practice Problems linked to LO 4
13.8 Initial and Final Value Theorems
Practice Problems linked to LO 5
Frequently Asked Questions linked to LO 5
Additional Solved Problems
Answers to Practice Problems
Objective-Type Questions
563
563
563
564
564
567
570
573
583
585
585
586
587
587
593
594
596
597
597
611
612
Chapter 14 Application of the Laplace Transform in Circuit Analysis
Learning Objectives
14.1 Circuit Elements in the S-Domain
Frequently Asked Questions linked to LO 1
14.2 Applications
Practice Problems linked to LO 2
Frequently Asked Questions linked to LO 2
14.3 Transfer Function
614
614
614
616
616
623
626
626
Contents
14.4 Use of Transfer Function in Circuit Analysis
Practice Problems linked to LO 3
Frequently Asked Questions linked to LO 3
14.5 The Transfer Function and the Convolution Integral
14.6 The Transfer Function and the Steady-State Sinusoidal Response
Practice Problems linked to LO 4
14.7 The Impulse Function in Circuit Analysis
Additional Solved Problems
PSpice Problems
Answers to Practice Problems
Objective-Type Questions
Chapter 15 S-Domain Analysis
Learning Objectives
15.1 Concept of Complex Frequency
15.2 Physical Interpretation of Complex Frequency
15.3 Transform Impedance and Transform Circuits
Frequently Asked Questions linked to LO 2
15.4 Series and Parallel Combinations of Elements
15.5 Terminal Pairs or Ports
15.6 Network Functions For One-Port and Two-Port Networks
Practice Problems linked to LO 4
Frequently Asked Questions linked to LO 4
15.7 Poles and Zeros of Network Functions
15.8 Significance of Poles and Zeros
Practice Problems linked to LO 5
Frequently Asked Questions linked to LO 5
15.9 Properties of Driving Point Functions
15.10 Properties of Transfer Functions
15.11 Necessary Conditions for a Driving Point Function
15.12 Necessary Conditions for Transfer Functions
Frequently Asked Questions linked to LO 6
15.13 Time-Domain Response From Pole Zero Plot
15.14 Amplitude and Phase Response From Pole Zero Plot
Practice Problems linked to LO 7
Frequently Asked Questions linked to LO 7
15.15 Stability Criterion for an Active Network
15.16 Routh Criteria
Practice Problems linked to LO 8
Frequently Asked Questions linked to LO 8
Additional Solved Problems
Answers to Practice Problems
Objective-Type Questions
xv
627
628
628
628
632
633
633
637
662
664
665
667
667
667
668
670
673
674
677
677
680
680
681
682
683
684
684
687
688
689
689
690
691
691
692
692
693
695
696
696
705
705
xvi
Contents
Chapter 16 Two-Port Networks
Learning Objectives
16.1 Two-Port Network
16.2 Open-Circuit Impedance (Z) Parameters
16.3 Short-Circuit Admittance (Y) Parameters
Practice Problems linked to LO 2
Frequently Asked Questions linked to LO 2
16.4 Transmission (ABCD) Parameters
16.5 Inverse Transmission (A9 B9 C9 D9) Parameters
Practice Problems linked to LO 3
Frequently Asked Questions linked to LO 4
16.6 Hybrid (H) Parameters
16.7 Inverse Hybrid (G) Parameters
Practice Problems linked to LO 4
Frequently Asked Questions linked to LO 4
16.8 Inter-Relationships of Different Parameters
Practice Problems linked to LO 5
Frequently Asked Questions linked to LO 5
16.9 Interconnection of Two-Port Networks
16.10 T- and p-Representations
Practice Problems linked to LO 7
Frequently Asked Questions linked to LO 7
16.11 Terminated Two-Port Network
16.12 Lattice Networks
16.13 Image Parameters
Practice Problems linked to LO 10
Frequently Asked Questions linked to LO 10
Additional Solved Problems
PSpice Problems
Answers to Practice Problems
Objective-Type Questions
707
707
707
708
711
714
717
719
722
722
723
724
726
727
728
729
733
734
734
737
740
741
741
747
750
754
754
754
775
778
779
Chapter 17 Filters and Attenuators
Learning Objectives
17.1 Classification of Filters
Frequently Asked Questions linked to LO 1
17.2 Filter Networks
17.3 Equations of Filter Networks
Frequently Asked Questions linked to LO 2
17.4 Classification of Pass Band and Stop Band
17.5 Characteristic Impedance in the Pass and Stop Bands
Practice Problems linked to LO 3
17.6 Constant-K Low-Pass Filter
17.7 Constant-K High Pass Filter
Practice Problems linked to LO 4
Frequently Asked Questions linked to LO 4
781
781
781
783
783
784
789
789
792
792
793
796
799
799
Contents
17.8 m-Derived T-Section
Practice Problems linked to LO 5
Frequently Asked Questions linked to LO 5
17.9 Band-Pass Filter
17.10 Band-Elimination Filter
Practice Problems linked to LO 6
Frequently Asked Questions linked to LO 6
17.11 Attenuators
17.12 T-Type Attenuator
17.13 -Type Attenuator
17.14 Lattice Attenuator
17.15 Bridged-T Attenuator
17.16 L-Type Attenuator
Practice Problems linked to LO 7
Frequently Asked Questions linked to LO 7
17.17 Equalisers
17.18 Inverse Network
17.19 Series Equaliser
17.20 Full-Series Equaliser
17.21 Shunt Equaliser
17.22 Full-Shunt Equaliser
17.23 Constant-Resistance Equaliser
17.24 Bridged-T Attenuation Equaliser
17.25 Bridged-T Phase Equaliser
17.26 Lattice-Attenuation Equaliser
17.27 Lattice-Phase Equaliser
Practice Problems linked to LO 8
Additional Solved Problems
PSpice Problems
Answers to Practice Problems
Objective-Type Questions
Chapter 18 Elements of Realizability and Synthesis of One-port Networks
Learning Objectives
18.1 Hurwitz Polynomials
Practice Problems linked to LO 1
Frequently Asked Questions linked to LO 1
18.2 Positive Real Functions
Practice Problems linked to LO 2
Frequently Asked Questions linked to LO 2
18.3 Frequency Response of Reactive One-Ports
18.4 Synthesis of Reactive One-Ports by Foster’s Method
18.5 Synthesis of Reactive One-Ports by the Cauer Method
Practice Problems linked to LO 4
Frequently Asked Questions linked to LO 4
18.6 Synthesis of R-L Network by the Foster Method
18.7 Synthesis of R-L Network by the Cauer Method
xvii
799
807
807
808
812
815
816
816
817
819
820
821
823
824
824
824
825
827
827
829
830
832
832
833
835
836
837
837
845
849
851
852
852
852
854
855
855
856
856
856
858
861
864
865
866
869
xviii
Contents
18.8 Synthesis of R-C Network by the Foster Method
18.9 Synthesis of R-C Network by the Cauer Method
Additional Solved Problems
Answers to Practice Problems
Objective-Type Questions
872
876
878
891
893
Chapter 19 An Introduction to PSpice
Learning Objectives
19.1 Introduction
19.2 What is Pspice?
19.3 Getting Started with Pspice
19.4 Simulation Steps
19.5 Component Values
19.6 DC Analysis and Control Statements
19.7 Dependent Sources
19.8 DC Sweep
19.9 AC Analysis and Control Statements
19.10 Transient Analysis
Additional Solved Problems
Test Your Skill in PSpice
895
895
895
895
896
896
897
897
900
903
904
905
908
916
Appendix A Fourier Series
Appendix B The j Factor
Index
918
928
931
Preface
This textbook is exclusively designed for electrical engineering students studying a basic level course in circuit
analysis offered to undergraduate students of various universities. This edition is prepared with students and
instructors in mind. The principal objectives of the book continue to be to provide an introduction to basic
concepts for circuit analysis, and to develop a strong foundation that can be used as the basis for further
study. To achieve these objectives, emphasis has been placed on basic laws, theorems, and techniques which
are used to develop a working knowledge of the methods of analysis, used in further topics of electrical
engineering. The mathematical complexity of the book remains at a level well within the grasp of college
first-year undergraduate students.
This book is designed for the third semester of EEE/ECE/EI/CSE students of various universities in the
country. This book enables the student have a firm grasp on the basic principles of Circuits and Networks:
Analysis and Synthesis. It lays emphasis on the basic laws, theorems, and techniques of analysis which helps
students develop the ability to design practical circuits that perform the desired operations.
The main objective of the revision was to align this extremely popular content with internationally approved
learning objectives for the course. Learning Objectives are the heart of every lesson, giving a purpose to
learning. They are the foundations for lesson planning so that the students have a sense of purpose to learning
and to know what is expected of them.
We received a great deal of useful feedback on the fourth edition, and we paid careful attention to it. While
attempting to improve the book in all dimensions, our main aim was increasing its usefulness to students
and instructors. The revised text and its underlying concepts & principles will be more interesting-to-read,
easy-to-understand, and logical-to-follow. Given here is a quick review of the principle newness of the fifth
edition over the fourth.
Each of the 19 chapters follows a common structure with a range of learning and assessment tools for
instructors and students.
In bringing out the fifth edition, we have taken advantage
of recent technological developments to create a wealth
of useful information not present in the physical book.
For students using smartphones and tablets, scanning QR
codes located within the chapters gives them immediate
access to more resources.
For interactive quiz with answers,
visit
http://qrcode.flipick.com/index.php/259
OR scan the QR code given here.
The QR code appearing at the last page of each chapter
gives students access to additional chapter resources which
include Interactive Quizzes.
xx
Preface
Learning Objectives
LEARNING OBJECTIVES
Each chapter begins with a list of key
Learning Objectives that are directly
tied to the chapter content. These help
in focussed planning for instructors and
methodical studying for students. The
chapters are now more modularised
this will help in systematic concept
development.
Arrangement of Pedagogy
The pedagogy is arranged as per levels of difficulty to enable the students to evaluate their learning levels.
This assessment of levels of difficulty is derived from Bloom’s taxonomy.
r indicates Level 1 and Level 2 i.e., Knowledge and
Comprehension based easy-to-solve problems
rr
indicates
Level 3 and Level 4 i.e., Application and
r
rr
Analysis based medium-difficulty problems
rrr
rrr indicates Level 5 and Level 6 i.e., Synthesis and
Evaluation based high-difficulty problems
Definitions and
Important Formulae
The average power is expressed in watts. It means the useful power transferred from the source to the
load, which is also called true power. If we consider a dc source applied to the network, true power is given
by the product of the voltage and the current. In case of sinusoidal voltage applied to the circuit, the product
of voltage and current is not the true power or average power. This product is called apparent power. The
apparent power is expressed in volt amperes, or simply VA.
∴
Apparent power 5 Veff Ieff
∴
Features like Definition and Important
Formulas are highlighted within
the text to draw special attention
to important concepts
In Eq. (6.10), the average power depends on the value of cos u; this is called the power factor of the circuit.
Pav
Power factor ( pf ) = cosu =
Veff I eff
Therefore, power factor is defined as the ratio of average power to the apparent power, whereas apparent
power is the product of the effective values of the current and the voltage. Power factor is also defined as the
factor with which the volt amperes are to be multiplied to get true power in the circuit.
In the case of sinusoidal sources, the power factor is the cosine of the phase angle between voltage and
current
pf 5 cos u
As the phase angle between voltage and total current increases, the power factor decreases. The smaller
the power factor, the smaller the power dissipation. The power factor varies from 0 to 1. For purely resistive
circuits, the phase angle between voltage and current is zero, and hence the power factor is unity. For
purely reactive circuits, the phase angle between voltage and current is 90°, and hence the power factor is
zero. In an RC circuit, the power factor is referred to as leading power factor because the current leads the
voltage. In an RL circuit, the power factor is referred to as lagging power factor because the current lags
behind the voltage.
xxi
Preface
Practice Problems linked to LO 2
Use step functions to write the expression for
the function shown in Fig. Q.1.
rrr13-2.2 Step functions can be used to define a window
function. Thus, u(t – 1) – u(t – 4) defines a
window 1 unit high and 3 units wide located
on the time axis between 1 and 4.
A function f (t) is defined as follows:
5
rr 13-2.1
Improved and
Expanded
In-text Exercises
EXAMPLE 2.7
This is by far the best
feature! Exam-friendly
pedagogy has been
arranged within the text and
linked after every Learning
Objective. This offers great
retention through looping
mechanism.
Write the mesh current equations in the circuit shown in Fig. 2.27,
and determine the currents.
Solution Assume two mesh currents in the direction as indicated
in Fig. 2.28.
The mesh current equations are
5I1 1 2(I1 – I2) 5 10
Fig. 2.27
Frequently Asked Questions linked to LO8
rr 9-8.1
r 9-8.2
rr 9-8.3
rr 9-8.4
A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW
when the line voltage is 220 V and the line current is 18.2 A. Determine the power factor of the
motor.
[AU Nov./Dec. 2012]
In a three-phase balanced delta system the voltage across R and Y is 400 0ºV. What will be the
voltage across Y and B? Assume RYB phase sequence.
[AU April/May 2011]
A balanced -connected load has one phase current IBC = 2 –90º A. Find the other phase current
and the three line currents if the system is an ABC system. If the line voltage is 100 V, what is the
load impedance?
[AU April/May 2011]
The power consumed in a three-phase, balanced star-connected load is 2 kW at a power factor of
0.8 lagging. The supply voltage is 400 V, 50 Hz. Calculate the resistance and reactance of each
phase.
[AU April/May 2011]
Additional Solved Problems
PROBLEM 9.1
The phase voltage of a star-connected three-phase ac generator is 230 V. Calculate the (a) line voltage,
(b) active power output if the line current of the system is 15 A at a power factor of 0.7, and (c) active and
reactive components of the phase currents.
Chapter-end Exercises
Solution

VPh 5
IL 5 IPh 5
f5
Pedagogy includes
Additional Solved Problems,
PSpice Problems, and
Objective Type Questions,
which is also integrated
through QR Codes are
featured at the
end of the chapter.
f5
PSpice Problems
PROBLEM 5.1
For the parallel circuit shown in Fig. 5.39, find the magnitude of current in each branch and the total current.
What is the phase angle between the applied voltage and total current.
Objective-Type Questions
r 9.1
3
r 9.2
A B C
A
B
C


2 
4 
(a ) Emax sin vt −  ; Emax sin vt − 


3 
3 


2 
4 
( b) Emax sin vt +  ; Emax sin vt + 


3 
3 
Emax
vt
Preface
xxii
The basic approach of the previous edition has been retained. All the elements with definitions, basic laws,
and configurations of the resistive circuits have been introduced in Chapter 1. Analysis of dc resistive circuits
using graph theory has been discussed in Chapter 2. Network theorems on resistive circuits have been
presented in Chapter 3. The concept of alternating currents and voltages has been introduced in Chapter 4.
Due emphasis has been laid on finding out the average and rms values of different waveforms. Chapters 5
and 6 introduce the complex impedance, and the concept of power and power factor respectively.
The steady-state analysis of ac circuits, including network theorems, has been discussed in Chapter 7. In all
the above chapters, problems, tutorials, and objective questions on dependant sources have been discussed.
Resonance phenomenon in series and parallel circuits, and locus diagrams are presented in Chapter 8.
A comprehensive study of polyphase systems and power measurement in both balanced and unbalanced
circuits is presented in Chapter 9. A brief study of coupled circuits, tuned circuits, and magnetic circuits
is introduced in Chapter 10. The transient behaviour of dc and ac circuits and their responses has been
discussed in Chapter 11. The Fourier methods of waveform analysis and their applications in circuit analysis
have been discussed in Chapter 12.
Laplace transforms and their applications are presented in Chapters 13 and 14. A brief account of S-domain
analysis is presented in Chapter 15. The parameters of two-port networks and their inter-relations have been
discussed in Chapter 16. Various types of basic filters, attenuators, and equalizers have been discussed in
Chapter 17. Elements of realizability and synthesis of one-port RL, RC networks have been briefly discussed
in Chapter 18. A chapter on introduction to PSpice has been included as Chapter 19. The book also includes
brief coverage of active filters and the j-operator as appendices.
The text is supported by an exhaustive website accessible at
http://highered.mheducation.com/sites/9339219600 with the following supplements:
- Solutions Manual
- PowerPoint Lecture Slides
- Solutions to Frequently Asked Questions
Many people have helped us produce this book. We extend our gratitude to them for assisting us in their
own individual ways to shape the book into the final form. We would like to express our sincere thanks to
the management of RVR & JC College of engineering, particularly to the President, Dr K Basava Punnaiah;
Secretary and Correspondent, Sri R Gopala Krishna; and Treasurer, Dr M Gopalkrishna. Our heartfelt
gratitude is due to the management of Sir C R Reddy College of Engineering, particularly the President,
Sri Alluri Indra Kumar; Secretary, Dr M B S V Prasad; Vice Presidents, Sri Maganti Venkateswara Rao and
Sri K Rajendra Vara Prasada Rao; Joint Secretaries, Sri M B V N Prasad and Sri Kodali Venkata Subba Rao;
Treasurer, Sri V V Raja Bhushan Prasad; and Correspondent, Sri Chitturi Janaki Ramayya for providing us
a conductive atmosphere.
Preface
xxiii
We are indebted to Dr G Sambasiva Rao, Principal of Sir C R Reddy College of Engineering, Prof. K A
Gopalarao of Andhra University; Sri B Amarendra Reddy of Andhra University; and Dr K K R of Gowtham
Concepts School for their support throughout the work. We are thankful to Prof. G S N Raju, Prof. M Ravindra
Reddy, Sri T Sreerama Murthy, and many other colleagues for their invaluable suggestions. We are extremely
appreciative of Prof. K Swarnasri who selflessly devoted her time in thoroughly reviewing this edition of the
book and implementing the PSpice Problems. We express thanks to Dr G Sambasiva Rao and Sri N C Kotaiah
for helping us in preparing solutions to the frequently asked questions.
We also thank the students of the ECE Department, particularly Ms V Krishna Geethika, Ms P Navya Sindhu
of RVR & JC College of Engineering and T Jayanth Kumar, T Sumanth Babu, Alapati Haritha, A Siva Kumar,
and P Alakananda who were involved directly or indirectly with the writing of this book. We acknowledge the
contribution and help of Mr Srivamsi Krishna Ponnekanti, Project Fellow of Major Research Project (UGC)
in the ECE department of RVR & JC College of Engineering. We are thankful to Mr D S R Anjaneyalu and
K Srinivas for the error-free typing of the manuscript.
We acknowledge the following reviewers who took out time to review the manuscript and send us useful
suggestions:
M S Naruka
IEC College of Engineering and Technology, Greater Noida, Uttar Pradesh
Gagan Deep Yadav
Yamuna Institute of Engineering and Technology, Lucknow, Uttar Pradesh
Kshipra Shingwekar
Jai Narain College of Engineering, Bhopal, Madhya Pradesh
Mukesh Kumawat
NRI Institute of Research and Technology, Bhopal, Madhya Pradesh
Amitava Biswas
Academy of Technology, Hooghly, West Bengal
Pravin Patil
Vidyalankar Institute of Technology, Mumbai, Maharashtra
Ashwini Kotrashetti
Don Bosco Institute of Technology, Mumbai, Maharashtra
Mahadik Shamala Rajaram Dr J J Magdum College of Engineering, Jaysingpur, Maharashtra
Walchand College of Engineering, Sangli, Maharashtra
P Balamurugan
Vellore Institute of Technology, Chennai
R Ganesan
Sethu Institute of Technology, Virudhunagar, Tamil Nadu
Vishwanath Hegde
Malnad College of Engineering, Hassan, Karnataka
V Vijaya Kumar Raju
Sri Vishnu Engineering College for Women (SVECW), Bhimavaram,
Andhra Pradesh
K Ramalingeswara Prasad Lakireddy Bali Reddy College of Engineering, Mylavaram, Andhra Pradesh
S V D Anil Kumar
St. Ann's College of Engineering and Technology, Hyderabad, Telangana
We wish to express our appreciation to the various members of McGraw Hill Education (India) who handled
the book at different stages.
Finally, we thank our family members—Madhavi, Aparna, A V Yashwanth, P Siddartha, and P Yudhister—
whose invaluable support made the whole project possible.
xxiv
Preface
Despite the best efforts put in by us and our team, it is possible that some unintentional errors might have
eluded us. We shall acknowledge with gratitude if any of these is pointed out. Any suggestions or comments
from the readers for improving future editions of the book may please be sent to the publisher’s email address.
McGraw Hill Education (India) invites suggestions and comments from you, all of which can be sent to
info.india@mheducation.com (kindly mention the title and author name in the subject line).
Piracy-related issues may also be reported.
1
LEARNING OBJECTIVES
1.1
VOLTAGE
According to the structure of an atom, we know that there are two types of
LO 1
charges: positive and negative. A force of attraction exists between these positive
and negative charges. A certain amount of energy (work) is required to overcome
the force and move the charges through a specific distance. All opposite charges
possess a certain amount of potential energy because of the separation between
them. The difference in potential energy of the charges is called the potential difference.
Potential difference in electrical terminology is known as voltage, and is denoted either by V or v. It is
expressed in terms of energy (W ) per unit charge (Q), i.e.,
W
dw
V=
or v =
Q
dq
dw is the small change in energy, and
dq is the small change in charge.
where energy (W ) is expressed in joules (J), charge (Q) in coulombs (C), and voltage (V ) in volts (V). One
volt is the potential difference between two points when one joule of energy is used to pass one coulomb
of charge from one point to the other.
2
Circuits and Networks
EXAMPLE 1.1
If 70 J of energy is available for every 30 C of charge, what is the voltage?
W 70
Solution V = = = 2.33 V
Q 30
1.2
CURRENT
LO 1
There are free electrons available in all semiconductive and
conductive materials. These free electrons move at random in
all directions within the structure in the absence of external
pressure or voltage. If a certain amount of voltage is applied
across the material, all the free electrons move in one direction
depending on the polarity of the applied voltage, as shown in
Fig. 1.1.
Fig. 1.1
This movement of electrons from one end of the material
to the other end constitutes an electric current, denoted by either I or i. The conventional direction of current
flow is opposite to the flow of 2ve charges, i.e. the electrons.
Current is defined as the rate of flow of electrons in a conductive or semiconductive material. It is measured
by the number of electrons that flow past a point in unit time. Expressed mathematically,
Q
I=
t
where I is the current, Q is the charge of electrons, and t is the time, or
dq
i=
dt
where dq is the small change in charge, and dt is the small change in time.
In practice, the unit ampere is used to measure current, denoted by A. One ampere is equal to one coulomb
per second. One coulomb is the charge carried by 6.25 3 1018 electrons. For example, an ordinary 80 W
domestic ceiling fan on 230 V supply takes a current of approximately 0.35 A. This means that electricity
is passing through the fan at the rate of 0.35 coulomb every second, i.e. 2.187 3 1018 electrons are passing
through the fan in every second; or simply, the current is 0.35 A.
EXAMPLE 1.2
Five coulombs of charge flow past a given point in a wire in 2 s. How many amperes of current is flowing?
Solution
1.3
I=
Q 5
= = 2.5 A
t
2
POWER AND ENERGY
Energy is the capacity for doing work, i.e. energy is nothing but stored work.
Energy may exist in many forms such as mechanical, chemical, electrical and
so on. Power is the rate of change of energy, and is denoted by either P or p. If a
certain amount of energy is used over a certain length of time, then
energy W
dw
Power( P ) =
or p =
=
time
t
dt
where dw is the change in energy and dt is the change in time.
LO 2
3
Circuit Elements and Kirchhoff’s Laws
dw dw dq
= ×
dt dq dt
5 v 3 i 5 vi W
Energy is measured in joules (J), time in seconds (s), and power in watts (W).
By definition, one watt is the amount of power generated when one joule of energy is consumed in
one second. Thus, the number of joules consumed in one second is always equal to the number of watts.
Amounts of power less than one watt are usually expressed in fraction of watts in the field of electronics;
viz. milliwatts (mW) and microwatts (mW). In the electrical field, kilowatts (kW) and megawatts (MW) are
common units. Radio and television stations also use large amounts of power to transmit signals.
We can also write
p=
EXAMPLE 1.3
What is the power in watts if energy equal to 50 J is used in 2.5 s?
energy 50
Solution P =
=
= 20 W
time
2.5
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 2*
rrr 1-2.1
rrr 1-2.2
rrr 1-2.3
rrr 1-2.4
rrr 1-2.5
rrr1-2.6
rrr 1-2.7
A resistor of 30 V has a voltage rating of 500 V; what is its power rating?
A 6.8 kV resistor has burned out in a circuit. It has to be replaced with another resistor with the
same ohmic value. If the resistor carries 10 mA, what should be its power rating?
A 12 V source is connected to a 10 V resistor.
(a) How much energy is used in two minutes?
(b) If the resistor is disconnected after one minute, does the power absorbed in the resistor
increase or decrease?
A capacitor is charged to 50 mC. The voltage across the capacitor is 150 V. It is then connected to
another capacitor four times the capacitance of the first capacitor. Find the loss of energy.
The current in the 5 V resistance of the circuit
shown in Fig. Q.5 is 5 A. Find the current in the
10 V resistor. Calculate the power consumed
by the 5 V resistor.
Find the power absorbed by the 5 V resistor
shown in Fig. Q.6.
Find the power absorbed by each circuit
Fig. Q.5
element of Fig. Q.7 it the control for the
dependent source is (a) 0.8 ix and (b) 0.8 iy.
Fig. Q.6
*Note:
rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
Fig. Q.7
4
Circuits and Networks
rrr 1-2.8
Find the power absorbed by each element and show that the algebraic sum of powers is zero in the
circuit shown in Fig. Q.8.
Fig. Q.8
rrr1-2.9
Find the power absorbed by the element X in the circuit shown
in Fig. Q.9 if it is a (a) 4 kV resistor, (b) 20 mA independent
current source, reference arrow directed towards right, and (c)
dependent current source, reference arrow directed towards right,
labeled 2ix.
Fig. Q.9
Frequently Asked Questions linked to LO 2*
rrr 1-2.1
1.4
A bulb is rated as 230 V, 230 W. Find the rated current and resistance of the filament.
[AU April/May 2011]
THE CIRCUIT
Simply put, an electric circuit consists of three parts: (1) energy source, such
LO 3
as battery or generator, (2) the load or sink, such as lamp or motor, and (3)
connecting wires as shown in Fig. 1.2. This arrangement represents a simple
circuit. A battery is connected to a lamp with two wires. The purpose of the
circuit is to transfer energy from source (battery) to the load (lamp). And this is
accomplished by the passage of electrons through wires around the circuit.
The current flows through the filament of the lamp, causing it to
emit visible light. The current flows through the battery by chemical
action. A closed circuit is defined as a circuit in which the current
has a complete path to flow. When the current path is broken so that
current cannot flow, the circuit is called an open circuit.
More specifically, interconnection of two or more simple circuit
elements (viz. voltage sources, resistors, inductors and capacitors) is
called an electric network. If a network contains at least one closed
path, it is called an electric circuit. By definition, a simple circuit
Fig. 1.2
element is the mathematical model of two terminal electrical devices,
and it can be completely characterised by its voltage and current. Evidently then, a physical circuit must
provide means for the transfer of energy.
Broadly, network elements may be classified into four groups, viz.,
1. Active or passive
2. Unilateral or bilateral
3. Linear or nonlinear
4. Lumped or distributed
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
Circuit Elements and Kirchhoff’s Laws
1.4.1
5
Active and Passive
Energy sources (voltage or current sources) are active elements, capable of delivering power to some external
device. Passive elements are those which are capable only of receiving power. Some passive elements like
inductors and capacitors are capable of storing a finite amount of energy, and return it later to an external
element. More specifically, an active element is capable of delivering an average power greater than zero to
some external device over an infinite time interval. For example, ideal sources are active elements. A passive
element is defined as one that cannot supply average power that is greater than zero over an infinite time
interval. Resistors, capacitors, and inductors fall into this category.
1.4.2 Bilateral and Unilateral
In the bilateral element, the voltage-current relation is the same for current flowing in either direction.
In contrast, a unilateral element has different relations between voltage and current for the two possible
directions of current. Examples of bilateral elements are elements made of high conductivity materials in
general. Vacuum diodes, silicon diodes, and metal rectifiers are examples of unilateral elements.
1.4.3 Linear and Nonlinear Elements
An element is said to be linear, if its voltage-current characteristic is at all times a straight line through the
origin. For example, the current passing through a resistor is proportional to the voltage applied through
it, and the relation is expressed as V  I or V 5 IR. A linear element or network is one which satisfies the
principle of superposition, i.e., the principle of homogeneity and additivity. An element which does not
satisfy the above principle is called a nonlinear element.
1.4.4 Lumped and Distributed
Lumped elements are those elements which are very small in size and in which simultaneous actions take
place for any given cause at the same instant of time. Typical lumped elements are capacitors, resistors,
inductors and transformers. Generally, the elements are considered as lumped when their size is very small
compared to the wave length of the applied signal. Distributed elements, on the other hand, are those which
are not electrically separable for analytical purposes. For example, a transmission line which has distributed
resistance, inductance and capacitance along its length may extend for hundreds of miles.
Frequently Asked Questions linked to LO 3
rrr 1-3.1
Explain the terms: (a) Linear (b) Bilateral (c) Passive (d) Lumped parameter. [GTU Dec. 2010]
Explain the terms: (a) Nonlinear (b) Unilateral (c) Passive (d) Lumped parameter. [GTU Dec. 2012]
rrr 1-3.3 What are the network elements? Explain them.
[JNTU Nov. 2012]
rrr 1-3.4 What do you mean by a linear bilateral network?
[PTU 2011-2012]
rrr 1-3.2
1.5
RESISTANCE PARAMETER
When a current flows in a material, the free electrons move through the material
LO 4
and collide with other atoms. These collisions cause the electrons to lose some
of their energy. This loss of energy per unit charge is the drop in potential across
the material. The amount of energy lost by the electrons is related to the physical property of the material.
These collisions restrict the movement of electrons. The property of a material to restrict the flow of electrons
is called resistance, denoted by R. The symbol for the resistor is shown in Fig. 1.3.
6
Circuits and Networks
The unit of resistance is ohm (V). Ohm is defined as the resistance offered
by the material when a current of one ampere flows between two terminals
Fig. 1.3
with one volt applied across it.
According to Ohm’s law, the current is directly proportional to the voltage and inversely proportional to
the total resistance of the circuit, i.e.,
V
I=
R
v
or i =
R
We can write the above equation in terms of charge as follows.
LO-4
dq
v
V = R , or i = = Gv
dt
R
where G is the conductance of a conductor. The units of resistance and conductance are ohm (V) and mho
(�) respectively.
When current flows through any resistive material, heat is generated by the collision of electrons with
other atomic particles. The power absorbed by the resistor is converted to heat. The power absorbed by the
resistor is given by
P 5 vi 5 (iR)i 5 i2 R
where i is the current in the resistor in amps, and v is the voltage across the resistor in volts. Energy lost in a
resistance in time t is given by
t
W = ∫ pdt = pt = i 2 Rt =
0
v2
t
R
where v is the volts,
R is in ohms,
t is in seconds, and
W is in joules.
EXAMPLE 1.4
A 10 V resistor is connected across a 12 V battery. How much current flows through the resistor?
Solution V 5 IR
I=
V 12
= = 1.2 A
R 10
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 4
A resistor with a current of 2 A through it converts 1000 J of electrical energy to heat energy in 15
s. What is the voltage across the resistor?
rrr 1-4.2 The filament of a light bulb in the circuit has a certain amount of resistance. If the bulb operates
with 120 V and 0.8 A of current, what is the resistance of its filament?
rrr 1-4.1
Frequently Asked Questions linked to LO 4
rrr 1-4.1
State the limitation of Ohm’s law.
[AU May/June 2013]
Circuit Elements and Kirchhoff’s Laws
1.6
7
INDUCTANCE PARAMETER
A wire of certain length, when twisted into a coil becomes a basic inductor. If
LO 5
current is made to pass through an inductor, an electromagnetic field is formed.
A change in the magnitude of the current changes the electromagnetic field.
Increase in current expands the fields, and decrease in current reduces it. Therefore, a change in current
produces change in the electromagnetic field, which induces a voltage across the coil according to Faraday’s
law of electromagnetic induction.
The unit of inductance is henry, denoted by H. By definition, the inductance is one henry when current
through the coil, changing at the rate of one ampere per second, induces one volt
across the coil. The symbol for inductance is shown in Fig. 1.4.
Fig. 1.4
The current-voltage relation is given by
di
v=L
dt
where v is the voltage across inductor in volts, and i is the current through inductor in amps. We can rewrite
the above equations as
di =
1
vdt
L
Integrating both sides, we get
t
t
1
di = ∫ vdt
L 0
∫
0
0
0t
t
i (t ) − i ( 0 ) =
1
vdt
L ∫0
t
i (t ) =
1
vdt + i(0)
L ∫0
From the above equation, we note that the current in an inductor is dependent upon the integral of the
voltage across its terminals and the initial current in the coil, i(0).
The power absorbed by the inductor is
P = vi = Li
di
watts
dt
The energy stored by the inductor is
t
W = ∫ pdt
0
t
= ∫ Li
0
di
Li 2
dt =
2
dt
From the above discussion, we can conclude the following:
8
Circuits and Networks
1. The induced voltage across an inductor is zero if the current through it is constant. That means an
inductor acts as short circuit to dc.
2. A small change in current within zero time through an inductor gives an infinite voltage across the
inductor, which is physically impossible. In a fixed inductor, the current cannot change abruptly.
3. The inductor can store finite amount of energy, even if the voltage across the inductor is zero, and
4. A pure inductor never dissipates energy, only stores it. That is why it is also called a non-dissipative
passive element. However, physical inductors dissipate power due to internal resistance.
EXAMPLE 1.5
The current in a 2 H inductor varies at a rate of 2 A/s. Find the voltage across the inductor and the energy
stored in the magnetic field after 2 s.
Solution
1.7
di
dt
= 2× 4 = 8 V
1
W = Li 2
2
1
= × 2×( 4) 2 = 16 J
2
v=L
CAPACITANCE PARAMETER
Any two conducting surfaces separated by an insulating medium exhibit the property
LO 6
of a capacitor. The conducting surfaces are called electrodes, and the insulating
medium is called dielectric. A capacitor stores energy in the form of an electric field
that is established by the opposite charges on the two electrodes. The electric field
is represented by lines of force between the positive and negative charges, and is concentrated within the
dielectric. The amount of charge per unit voltage that is capacitor can store is its capacitance, denoted by
C. The unit of capacitance is Farad denoted by F. By definition, one Farad is
the amount of capacitance when one coulomb of charge is stored with one volt
across the plates. The symbol for capacitance is shown in Fig. 1.5.
Fig. 1.5
A capacitor is said to have greater capacitance if it can store more charge per
unit voltage and the capacitance is given by
Q
q
, or C =
V
v
(lowercase letters stress instantaneous values)
We can write the above equation in terms of current as
C=
i =C
dv
dt


∵ i = dq 

dt 
where v is the voltage across capacitor and i is the current through it.
1
dv = idt
C
Circuit Elements and Kirchhoff’s Laws
9
Integrating both sides, we have
t
t
1
C
∫ idt
1
v (t ) − v ( 0 ) =
C
∫ idt
∫
0
dv =
1
v (t ) =
C
0
t
0
t
∫ idt + v(0)
0
where v (0) indicates the initial voltage across the capacitor.
From the above equation, the voltage in a capacitor is dependent upon the integral of the current through
it, and the initial voltage across it.
The power absorbed by the capacitor is given by
dv
p = vi = vC
dt
The energy stored by the capacitor is
t
t
W = ∫ pdt = ∫ v C
0
0
dv
dt
dt
1
W = Cv 2
2
From the above discussion, we can conclude the following:
1. The current in a capacitor is zero if the voltage across it is constant; that means, the capacitor acts as
an open circuit to dc
2. A small change in voltage across a capacitance within zero time gives an infinite current through the
capacitor, which is physically impossible. In a fixed capacitance the voltage cannot change abruptly.
3. The capacitor can store a finite amount of energy, even if the current through it is zero, and
4. A pure capacitor never dissipates energy, but only stores it; that is why it is called non-dissipative
passive element. However, physical capacitors dissipate power due to internal resistance.
EXAMPLE 1.6
A capacitor having a capacitance 2 mF is charged to a voltage of 1000 V. Calculate the stored energy in joules.
1
1
Solution W = Cv 2 = × 2×10−6 ×(1000) 2 = 1 J
2
2
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 6
rrr 1-6.1
(a) Determine the current in each of the following cases:
(i) 75 C in 1 s
(ii) 10 C in 0.5 s
(iii) 5 C in 2 s
(b) How long does it take 10 C to flow past a point if the current is 5 A?
rrr 1-6.2 Find the capacitance of a circuit in which an applied voltage of 20 V gives an energy store
of 0.3 J.
rrr 1-6.3 The voltage across two parallel capacitors of 5 mF and 3 mF changes uniformly from 30 to 75 V
in 10 ms. Calculate the rate of change of voltage for (a) each capacitor, and (b) the combination.
10
Circuits and Networks
Frequently Asked Questions linked to LO 6
rrr1-6.1
When a dc voltage is applied to a capacitor, voltage across its terminals is found to build up in
accordance with vc = 50(1 – e–100t). After 0.01 s, the current flow is equal to 2 mA.
(a) Find the value of capacitance in farad.
(b) How much energy is stored in the electric field?
[AU May/June 2014]
rrr 1-6.2 Give the relation between energy (E) and power (P). Derive the equations for the energy stored in a
capacitor (C) and an inductor (L) using P = VI
[GTU May 2011]
1.8
ENERGY SOURCES
According to their terminal voltage–current characteristics, electrical energy sources
LO 7
are categorised into ideal voltage sources and ideal current sources. Further they can
be divided into independent and dependent sources.
An ideal voltage source is a two-terminal element in which the voltage vs is
completely independent of the current is through its terminals. The representation of ideal constant voltage
source is shown in Fig. 1.6 (a).
Fig. 1.6
If we observe the v–i characteristics for an ideal voltage source as shown in Fig. 1.6 (c) at any time, the
value of the terminal voltage vs is constant with respect to the value of current is. Whenever vs 5 0, the voltage
source is the same as that of a short circuit. Voltage sources need not have constant magnitude; in many
cases the specified voltage may be time-dependent like a
sinusoidal waveform. This may be represented as shown in
Fig. 1.6 (b). In many practical voltage sources, the internal
resistance is represented in series with the source as shown
in Fig. 1.7 (a). In this, the voltage across the terminals falls
as the current through it increases, as shown in Fig. 1.7 (b).
The terminal voltage vt depends on the source current as
shown in Fig. 1.7 (b), where vt 5 vs – is R.
Fig. 1.7
An ideal constant current source is a two-terminal element in
which the current is completely independent of the
LO-7
voltage vs across its terminals. Like voltage sources
we can have current sources of constant magnitude is
or sources whose current varies with time is(t). The
representation of an ideal current source is shown in
Fig. 1.8 (a).
If we observe the v – i characteristics for an
ideal current source as shown in Fig. 1.8 (b), at
Fig. 1.8
11
Circuit Elements and Kirchhoff’s Laws
any time the value of the current is is constant with respect to
the voltage across it. In many practical current sources, the
resistance is in parallel with a source as shown in Fig. 1.9 (a).
In this the magnitude of the current falls as the voltage across
its terminals increases. Its terminal v – i characteristic is
shown in Fig. 1.9 (b). The terminal current is given by it 5 is
– (vs/R), where R is the internal resistance of the ideal current
source.
Fig. 1.9
The two types of ideal sources we have discussed are
independent sources for which voltage and current are independent and are not affected by other parts of the
circuit. In the case of dependent sources, the source voltage or current is not fixed, but is dependent on the
voltage or current existing at some other location in the circuit.
Dependent or controlled sources are of the following types:
1. voltage controlled voltage source (VCVS)
2. current controlled voltage source (CCVS)
3. voltage controlled current source (VCCS)
4. current controlled current source (CCCS)
These are represented in a circuit diagram by the symbol shown
in Fig. 1.10. These types of sources mainly occur in the analysis of
Fig. 1.10
equivalent circuits of transistors.
Frequently Asked Questions linked to LO 7
rrr 1-7.1
Explain about voltage source and current source. Include ideal, practical, independent and dependent
sources in your explanation.
[GTU Dec. 2010]
rrr 1-7.2 Explain following in brief: ideal and practical energy sources.
[GTU Dec. 2010]
rrr 1-7.3 What are the types of sources? Explain them with suitable diagrams and characteristics.
[JNTU Nov. 2012]
1.9
KIRCHHOFF’S VOLTAGE LAW
Kirchhoff’s voltage law states that the algebraic sum of all branch voltages
LO 8
around any closed path in a circuit is always zero at all instants of time. When
the current passes through a resistor, there is a loss of energy and, therefore, a
voltage drop. In any element, the current always flows from higher potential to
lower potential. Consider the circuit in Fig. 1.11. It is customary to take the direction of current I as indicated in
the figure, i.e. it leaves the positive terminal of the voltage
source and enters into the negative terminal.
As the current passes through the circuit, the sum of the
voltage drop around the loop is equal to the total voltage in
that loop. Here, the polarities are attributed to the resistors
to indicate that the voltages at points a, c, and e are more
than the voltages at b, d, and f, respectively, as the current
passes from a to f.
Vs 5 V1 1 V2 1 V3
Fig. 1.11
12
Circuits and Networks
Fig. 1.12
Fig. 1.13
Consider the problem of finding out the current supplied by the source V in the circuit shown in Fig. 1.12.
Our first step is to assume the reference current direction and to indicate the polarities for different
elements. (See Fig. 1.13).
By using Ohm’s law, we find the voltage across each resistor as follows.
VR1 5 IR1, VR2 5 IR2, VR3 5 IR3
where VR1, VR2 and VR3 are the voltages across R1, R2 and R3, respectively. Finally, by applying Kirchhoff’s
law, we can form the equations
V 5 VR1 1 VR2 1 VR3
V 5 IR1 1 IR2 1 IR3
From the above equation, the current delivered by the source is given by
I=
V
R1 + R2 + R3
EXAMPLE 1.7
For the circuit shown in Fig. 1.14, determine the unknown voltage drop V1.
Fig. 1.14
Solution According to Kirchhoff’s voltage law, the sum of the potential drops is equal to the sum of the
potential rises.
Therefore, 30 5 2 1 1 1 V1 1 3 1 5
or
V1 5 30 – 11 5 19 V
13
Circuit Elements and Kirchhoff’s Laws
EXAMPLE 1.8
What is the current in the circuit shown in Fig. 1.15? Determine the voltage across each resistor.
Fig. 1.15
Solution We assume the current I in the clockwise direction and indicate polarities (Fig. 1.16). By using
Ohm’s law, we find the voltage drops across each resistor.
VIM 5 I, V3.1M 5 3.1 I
V500K 5 0.5 I, V400K 5 0.4 I
Fig. 1.16
Now, by applying Kirchhoff’s voltage law, we form the equation
10 5 I 1 3.1 I 1 0.5 I 1 0.4 I
or 5 I 5 10
or I 5 2 mA
voltage across each resistor is as follows:
V1M 5 1 3 2 5 2.0 V
V3.1M 5 3.1 3 2 5 6.2 V
V400K 5 0.4 3 2 5 0.8 V
V500K 5 0.5 3 2 5 1.0 V
EXAMPLE 1.9
In the circuit given in Fig. 1.17, find (a) the current I, and (b) the voltage across 30 V.
Fig. 1.17
Solution We redraw the circuit as shown in Fig. 1.18 and assume
current direction and indicate the assumed polarities of resistors.
By using Ohm’s law, we determine the voltage across each
resistor as
V8 5 8I, V30 5 30I, V2 5 2I
Fig. 1.18
14
Circuits and Networks
By applying Kirchhoff’s law, we get
100 5 8I 1 40 1 30I 1 2I
60
40 I = 60 or I = =1.5 A
40
voltage drop across 30 V 5 V30 5 30 3 1.5 5 45 V
1.10
VOLTAGE DIVISION
The series circuit acts as a voltage divider. Since the same current flows through each resistor, the voltage
drops are proportional to the values of resistors. Using this principle, different voltages can be obtained from
a single source, called a voltage divider. For example, the voltage across a 40 V resistor is twice that of 20 V
in a series circuit shown in Fig. 1.19.
Fig. 1.19
Fig. 1.20
In general, if the circuit consists of a number of series resistors, the total current is given by the total
voltage divided by equivalent resistance. This is shown in Fig. 1.20.
The current in the circuit is given by I 5 Vs/(R1 1 R2 1 … 1 Rm). The voltage across any resistor is
nothing but the current passing through it, multiplied by that particular resistor.
Therefore, VR1 5 IR1
VR2 5 IR2
VR3 5 IR3
�
VRm 5 IRm
or VRm =
Vs ( Rm )
R1 + R2 +…+ Rm
From the above equation, we can say that the voltage drop across any resistor, or a combination of resistors
in a series circuit is equal to the ratio of that resistance value to the total resistance, multiplied by the source
voltage, i.e.,
Vm =
Rm
Vs
RT
where Vm is the voltage across mth resistor, Rm is the resistance across which the voltage is to be determined
and RT is the total series resistance.
15
Circuit Elements and Kirchhoff’s Laws
EXAMPLE 1.10
What is the voltage across the 10 V resistor in Fig. 1.21.
Solution Voltage across 10 V =V = 50× 10 = 500 = 33.3 V
10
10 + 5
15
Fig. 1.21
EXAMPLE 1.11
Find the voltage between A and B in a voltage divider network shown in
Fig. 1.22.
Solution
1.11
9
Voltage across 9 kV =V9 =VAB =100× = 90 V
10
POWER IN A SERIES CIRCUIT
Fig. 1.22
LO 8
The total power supplied by the source in any series resistive circuit is equal to the sum of the powers in each
resistor in series, i.e.,
Ps 5 P1 1 P2 1 P3 1 … 1 Pm
where m is the number of resistors in series, PS is the total power supplied by source, and Pm is the power in
the last resistor in series. The total power in the series circuit is the total voltage applied to a circuit, multiplied
by the total current. Expressed mathematically,
V2
Ps =Vs I = I 2 RT = s
RT
where Vs is the total voltage applied, RT is the total resistance, and I is the total current.
EXAMPLE 1.12
Determine the total amount of power in the series circuit in Fig. 1.23.
Fig. 1.23
Solution Total resistance 5 5 1 2 1 1 1 25 10 V
16
Circuits and Networks
Vs2 (50) 2
=
= 250 W
10
RT
We find the power absorbed by each resistor.
50
Current =
=5A
10
We know
Ps =
P5 5 (5)2 3 5 5 125 W
P2 5 (5)2 3 2 5 50 W
P1 5 (5)2 3 1 5 25 W
P2 5 (5)2 3 2 5 50 W
The sum of these powers gives the total power supplied by the source PS 5 250 W.
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 8
rrr 1-8.1
What is the voltage VAB across the resistor shown in Fig. Q.1?
Fig. Q.1
rrr 1-8.2
The source voltage in the circuit shown in Fig. Q.2 is 100 V. How much voltage does each metre
read?
rrr 1-8.3 Find the node voltages for the network shown in Fig. Q.3 for the case when k 5 – 2 using PSpice.
Fig. Q.2
Fig. Q.3
If you wish to increase the amount of current in a resistor from 100 mA to 150 mA by changing
the 20 V source, by how many volts should you change the source? To what new value should you
set it?
rrr 1-8.5 The following voltage drops are measured across each of three resistors in series: 5.5 V, 7.2 V, and
12.3 V. What is the value of the source voltage to which these resistors are connected? A fourth
resistor is added to the circuit with a source voltage of 30 V. What should be the drop across the
fourth resistor?
rrr1-8.4
17
Circuit Elements and Kirchhoff’s Laws
Frequently Asked Questions linked to LO 8
rrr 1-8.1
rrr 1-8.2
State and explain Kirchhoff’s law.
[AU May/June 2013]
Find the current I and voltage across 30 Ohms of the circuit shown in Fig. Q.2. [AU May/June 2013]
rrr 1-8.3 Determine current in the circuit shown in Fig. Q.3.
[AU May/June 2014]
40V
Fig. Q.3
Fig. Q.2
rrr 1-8.4
Determine the current through the resistances in the bridge network shown in Fig. Q.4 using
Kirchhoff’s laws.
[AU May/June 2014]
Fig. Q.4
2W
rrr 1-8.5
State the voltage division principle for two resistors in series
and the current division principle for two resistors in parallel.
[AU May/June 2013]
rrr 1-8.6 Find the equivalent resistance of the circuit shown in Fig.
Q.6.
[AU May/June 2014]
1.12
KIRCHHOFF’S CURRENT LAW
2W
RT
2W
1.2 W
1W
Fig. Q.6
Kirchhoff’s current law states that the sum of the currents entering into any
LO 9
node is equal to the sum of the currents leaving that node. The node may be
an interconnection of two or more branches. In any parallel circuit, the node
LO-9
is a junction point of two or more branches. The total current entering into
a node is equal to the current leaving that node. For example, consider the
circuit shown in Fig. 1.24, which contains two nodes A and B. The total current IT entering node A is divided
into I1, I2 and I3. These currents flow out of the node A. According to Kirchhoff’s current law, the current
into node A is equal to the total current out of the node A: that is,
IT 5 I1 1 I2 1 I3. If we consider the node B, all three currents I1,
I2, I3 are entering B, and the total current IT is leaving the node B,
Kirchhoff’s current law formula at this node is therefore the same
as at the node A.
I1 1 I2 1 I3 5 IT
In general, the sum of the currents entering any point or node
or junction equal to sum of the currents leaving from that point or
Fig. 1.24
18
Circuits and Networks
node or junction as shown in Fig. 1.25.
I1 1 I2 1 I4 1 I7 5 I3 1 I5 1 I6
If all of the terms on the right side are brought over to the left side, their
signs change to negative and a zero is left on the right side, i.e.
I1 1 I2 1 I4 1 I7 – I3 – I5 – I6 5 0
This means that the algebraic sum of all the currents meeting at a junction
is equal to zero.
Fig. 1.25
EXAMPLE 1.13
Determine the current in all resistors in the circuit
shown in Fig. 1.26.
Solution The above circuit contains a single node ‘A’
with the reference node ‘B’. Our first step is to assume
the voltage V at the node A. In a parallel circuit, the same
voltage is applied across each element. According to
Ohm’s law, the currents passing through each element
are I1 5 V/2, I2 5 V/1, I3 5 V/5.
By applying Kirchhoff’s current law, we have
I = I1 + I 2 + I 3
I=
Fig. 1.26
V V V
+ +
2 1 5
 1 1 1
50 =V  + +  =V [ 0.5 +1+ 0.2 ]
 2 1 5 
50 500
=
= 29.41 V
1.7 17
Once we know the voltage V at the node A, we can find the current in any element by using Ohm’s law.
The current in the 2 V resistor is I1 5 29.41/2 5 14.705 A.
V V
Similarly I 2 =
− = 29.41 A
R2 1
V=
29.41
= 5.882 A
5
I1 = 14.7 A, I 2 = 29.4 A, and I 3 = 5.88 A
I3 =
∴
EXAMPLE 1.14
For the circuit shown in Fig. 1.27, find the voltage
across the 10 V resistor and the current passing
through it.
Fig. 1.27
Circuit Elements and Kirchhoff’s Laws
19
Solution The circuit shown above is a parallel circuit, and consists of a single node A. By assuming voltage
V at the node A w.r.t. B, we can find out the current in the 10 V branch (See Fig. 1.28).
Fig. 1.28
According to Kirchhoff’s current law,
I1 1 I2 1 I3 1 I4 1 5 5 10
By using Ohm’s law, we have
V
V
V
V
I1 = , I 2 = , I 3 = , I 4 =
5
10
2
1
V V V
+ + +V + 5 =10
5 10 2
1 1 1 
V  + + +1 = 5
 5 10 2 
V [ 0.2 + 0.1+ 0.5 +1]= 5
5
= 2.78 V
1.8
the voltage across the 10 V resistor is 2.78 V and the current passing through it is
V 2.78
I2 = =
= 0.278 A
10 10
V=
EXAMPLE 1.15
Determine the current through the resistance R3 in the circuit shown in Fig. 1.29.
Fig. 1.29
Solution According to Kirchhoff’s current law,
IT 5 I1 1 I2 1 I3
where IT is the total current and I1, I2, and I3 are the currents in resistances R1, R2 and R3 respectively.
50 5 30 1 10 1 I3
or I3 5 10 mA
20
Circuits and Networks
1.13
PARALLEL RESISTANCE
LO 9
When the circuit is connected in parallel, the total resistance of the circuit decreases as the number of resistors
connected in parallel increases. If we consider m parallel branches in a circuit as shown in Fig. 1.30, the
current equation is
IT 5 I1 1 I2 1 ... 1 Im
The same voltage is applied across each resistor. By applying Ohm’s law, the current in each branch is
given by
I1 =
Vs
V
V
, I 2 = s ,… I m = s
R1
R2
Rm
According to Kirchhoff’s current law,
IT 5 I1 1 I2 1 I3 1 ... 1 Im
Vs Vs Vs Vs
V
= + + +…+ s
RT R1 R2 R3
Rm
Fig. 1.30
From the above equation, we have
1
1
1
1
= + +…+
RT R1 R2
Rm
EXAMPLE 1.16
Determine the parallel resistance between points A and B of the circuit shown in Fig. 1.31.
Fig. 1.31
Solution
1
1
1
1
1
= + + +
RT R1 R2 R3 R4
1
1
1
1
1
= + + +
RT 10 20 30 40
5 0.1 1 0.05 1 0.033 1 0.025 5 0.208
or RT 5 4.8 V
21
Circuit Elements and Kirchhoff’s Laws
1.14
CURRENT DIVISION
In a parallel circuit, the current divides in all branches. Thus, a parallel
circuit acts as a current divider. The total current entering into the
parallel branches is divided into the branches currents according to the
resistance values. The branch having higher resistance allows lesser
current, and the branch with lower resistance allows more current. Let
us find the current division in the parallel circuit shown in Fig. 1.32.
The voltage applied across each resistor is Vs. The current passing
through each resistor is given by
V
V
I1 = s , I 2 = s
R1
R2
If RT is the total resistance, which is given by R1R2/(R1 1 R2),
V
V
Total current IT = s = s ( R1 + R2 )
RT R1 R2
or
IT =
LO 9
Fig. 1.32
I1 R1
( R1 + R2 ) since Vs = I1 R1
R1 R2
I1 = IT ⋅
R2
R1 + R2
R1
R1 + R2
From the above equations, we can conclude that the current in any branch is equal to the ratio of the
opposite branch resistance to the total resistance value, multiplied by the total current in the circuit. In general,
if the circuit consists of m branches, the current in any branch can be determined by
RT
Ii =
IT
Ri + RT
where Ii represents the current in the ith branch,
Ri is the resistance in the ith branch,
RT is the total parallel resistance to the ith branch, and
IT is the total current entering the circuit.
Similarly, I 2 = IT ⋅
EXAMPLE 1.17
Determine the current through each resistor in the circuit
shown in Fig. 1.33.
RT
Solution I1 = I T ×
( R1 + RT )
where
∴
RT =
R2 R3
=2 V
R2 + R3
R1 = 4 V
IT = 12 A
I1 = 12×
2
=4 A
2+4
Fig. 1.33
22
Circuits and Networks
2
=4 A
2+4
2
=4 A
and I 3 = 12×
2+4
Since all parallel branches have equal values of resistance, they share current equally.
Similarly, I 2 = 12×
1.15
POWER IN A PARALLEL CIRCUIT
LO 9
The total power supplied by the source in any parallel resistive circuit is equal to the sum of the powers in
each resistor in parallel, i.e.,
Ps 5 P1 1 P2 1 P3 1 ... 1 Pm
where m is the number of resistors in parallel, Ps is the total power, and Pm is the power in the last resistor.
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 9
rrr1-9.1
Using Ohm’s law and Kirchhoff’s laws on the circuit given in Fig. Q.1, find all the voltages and
currents.
Fig. Q.1
rrr 1-9.2
Using PSpice, solve for the current Iab in the following
circuit. (Fig. Q.2)
rrr1-9.3
Find Req for the resistive network shown in Fig. Q.3.
Fig. Q.3
Fig. Q.2
23
Circuit Elements and Kirchhoff’s Laws
rrr 1-9.4
For the circuit shown in Fig. Q.4, find the total resistance.
Fig. Q.4
rrr 1-9.5
In the network shown in Fig. Q.5, (a) let R 5 80 V,
find Req; (b) find R if Req 5 80 V; (c) find R if
R 5 Req.
rrr 1-9.6
Using the current divider formula, determine the
current in each branch of the circuit shown in Fig.
Q.6.
Fig. Q.5
Fig. Q.6
rrr1-9.7
Six lightbulbs are connected in parallel across 110 V. Each bulb is rated at 75 W. How much
current flows through each bulb, and what is the total current?
Frequently Asked Questions linked to LO 9
rrr 1-9.1
rrr 1-9.2
State and explain Kirchhoff’s laws.
[AU May/June 2013]
Determine the voltage across the 20 ohm resistor of the network shown in Fig. Q.2.
[AU May/June 2014]
8W
5 V –+
10 W
20 W
12 W
Fig. Q.2
rrr 1-9.3
Determine the current through the 20 V source in the circuit of
shown in Fig. Q.3.
[AU May/June 2014]
20 V –+
10 W
Fig. Q.3
1A
24
Circuits and Networks
rrr 1-9.4
State and explain Kirchhoff’s laws, with an example.
[AU April/May 2011]
State and explain Kirchhoff’s laws.
[JNTU Nov. 2012]
rrr 1-9.6 Find the equivalent conductance Geq.of the circuit shown in Fig. Q.6.
[AU Nov./Dec. 2012]
rrr 1-9.7 What is the magnitude of current drained from the 10 V source in the circuit shown in Fig. Q.7.
[JNTU Nov. 2012]
rrr 1-9.5
5S
Geq
6S
1W
P
2W
8S
3W
12 S
5W
2W
3W
10 V
Q
Fig. Q.6
Fig. Q.7
rrr 1-9.8
Three loads A,B, and C are connected in parallel to a 240 V source. Load A takes 9.6 kW, Load B
takes 60 A and load C has a resistance of 4.8 Ohms. Calculate (a) RA and RB, (b) the total current
(c) the total power, and (d) equivalent resistance.
[AU May/June 2013]
rrr 1-9.9 Determine the current in all the resistors of the circuit shown in Fig. Q.9.
[AU May/June 2014]
rrr 1-9.10 Determine the current through each resistor in the circuit shown in Fig. Q.10. [AU May/June 2014]
A
50 A
2A
i1
i2
i3
2W
1W
5W
Vs
4W
4W
4W
I1
I2
I3
B
Fig. Q.9
Fig. Q.10
rrr 1-9.11 Determine the current through the 2 W resistor in the circuit of Fig. Q.11.
rrr1-9.12
[AU May/June 2014]
Determine the current delivered by the source in the circuit shown in Fig. Q.12. [AU April/May 2011]
2W
1W
A
1W
+
5A
2W
30 V
1W
2W
2W
–
2W
4W
1W
C
2W
1W
2W
D
B
Fig. Q.11
Fig. Q.12
rrr 1-9.13 Calculate
the voltage that is to be connected across terminals x-y as shown in Fig. Q. 13 such that
the voltage across the 2 W resister is 10 V. Also find Ia and Ib. what is the total power loss in the
circuit?
[JNTU Nov. 2012]
Ia 4 W
x
2W
5W
Ib 6 W
y
Fig. Q.13
rrr 1-9.14 A resistance of 10 ohms is connected across a supply of 200 V. If a resistances R is now connected
in parallel with a 10-ohm resistance, the current draw from the supply gets doubled. Find the value
of unknown resistance.
[PTU 2011-12]
25
Circuit Elements and Kirchhoff’s Laws
Additional Solved Problems
PROBLEM 1.1
For the circuit shown in Fig. 1.34, find the total resistance
between terminals A and B, the total current drawn from a 6 V
source connected from A to B, and the current through 4.7 kV;
voltage across 3 kV.
4 kW
3 kW
10 kW
5 kW
Solution The circuit in Fig. 1.34 can be redrawn as shown in
4.7 kW
Fig. 1.34
Fig. 1.35 below.
From Fig. 1.35, the total resistance is
RT 5 10 k || 3 k || [4 k 1 5 k || 4.7 k]
10 k
3 kW
5 1.7 kV
(5 k//4.7 k)
Total current drawn by the circuit is
IT =
Fig. 1.35
6V
= 3.53 mA
1.7 kV
The current in the 10 kV resistor is
6V
∴ I10 k =
= 0.6 mA
10 k
The current in the 3 kV resistor is
6V
I 3k =
= 2 mA
3k Ω
The remaining current blows through the 4 kV resistor and the parallel combination of (5 kV||4.7 kV).
I4k 5 3.53 mA 2 2.6 mA 5 0.93 mA
The current in the 4.7 kV resistor is
I 4.7 k = 0.93×
5
= 0.47 mA
5 + 4.7
The voltage across the 3 kV resistor is
V3k 5 I3k R 5 2 3 1023 3 3 3 103 5 6 V
PROBLEM 1.2
A battery of unknown emf is connected across resistances as shown in Fig. 1.36. The voltage drop across the
8 V resistor is 20 V. What will be the current reading in the ammeter? What is the emf of the battery?
26
Circuits and Networks
Fig. 1.36
Solution From the circuit shown in Fig. 1.36, the current passing through the 8 V resistor is
20 V
= 2.5 A
8V
The current passing through the 15 V resistor is same as the ammeter reading.
the current passing through the 15 V resistor is
2.5×11
I15 =
= 0.71 A
11 + 28
Reading of the ammeter 5 0.71 A
The voltage across the 28 V (13 V in series with 15 V) resistor is
V28 5 0.71 3 28 5 19.88 volts
The voltage across the series arm (8 V in series with 11 V) resistor is
V19 5 2.5 3 19 5 47.5 volts
The emf of the battery 5 19.88 1 47.5 5 67.38 volts
I8 =
PROBLEM 1.3
An electric circuit has three terminals A,B,C. Between A and B is connected a 2 V resistor, between B and C
are connected a 7 V resistor and a 5 V resistor in parallel, and between A and C is connected a 1 V resistor.
A battery of 10 V is then connected between terminals A and C calculate (a) total current drawn from the
battery, (b) voltage across the 2 V resistor, and (c) current passing through the 5 V resistor.
Solution The circuit can be drawn as shown in Fig. 1.37 below.
The current passing through the 1 V resistor is
10
I1V = = 10 A
The current passing through the series parallel branch between
terminals A and C is
10
I2V =
=2A
2 + (7�5)
Fig. 1.37
Total current drawn from the battery is IT 5 10 1 2 5 12 A
Voltage across the 2 V resistor is V2 V 5 2 3 2 5 4 volts
The current passing though the 5 V resistor is
2× 7
I5V =
= 1.17 A
5+7
27
Circuit Elements and Kirchhoff’s Laws
PROBLEM 1.4
Using Ohm’s law and Kirchhoff’s laws on the circuit given in
Fig. 1.38, find Vin, Vs, and the power provided by the dependent
source.
Solution From the circuit shown in Fig. 1.38, applying
Kirchhoff’s current law, we have
i4V 5 i3V 1 i2V
Fig. 1.38
The current passing through the 4 V branch is i4 5 2 1 6 5 8 A
The voltage across the dependent source is
V4i4 = ( 4i4 ) = 4 ×8 = 32 V
The voltage across the 2 V resistor is V2V 5 6 3 2 5 12 V
the voltage across each branch is
V = V4i4 −V2 = 32 −12 = 20 V
The voltage across the 4 V branch is
V4 5 4 3 i4 5 4 3 8
5 32 V
According to Kirchhoff’s voltage law,
V −V4 + VS = 0
∴ 20 − 32 + VS = 0
VS = 12 V
Similarly, Kirchhoff’s voltage law is applied to the 3 V branch.
V 2 30 1 V3 2 Vin 5 0
20 2 30 1 6 2 Vin 5 0
From the above equation, the voltage
Vin 5 24 V
The power provided by the dependent source is
P4i4 = V4i4 ×i4i4
P4i4 = 4 ×i4 ×6 = 4 ×8×6
= 192 watts
PROBLEM 1.5
Find the power absorbed by each element in the circuit shown in Fig. 1.39.
1 kW
20 mA
4 kW
Fig. 1.39
28
Circuits and Networks
Solution The circuit shown in Fig. 1.39 can be redrawn as shown in Fig. 1.40.
4 kW
1 kW
Fig. 1.40
Applying Kirchhoff’s current in the single node circuit shown in Fig. 1.40, we have
V
3i1 + 20 ×10−3 −
+ i1 = 0
4 ×103
i1 = −V ×10−3
∴ − 3V ×10−3 + 20 ×10−3 − 0.25×10−3 V −V ×10−3 = 0
−4.25×10−3V = −20 ×10−3
V = 4.71 volts
The power absorbed by the 3i1 dependent current source is
P3i1 = (3i1 ) V = 3V ×10−3 ×V
= +66.55 mW
The power absorbed by the 20 mA current source is
P20 5 20 3 1023 3 V 5 294.2 mW
The power absorbed by the 4 kV resistor is
4.71× 4.71
P4 k =
= 5.55 mW
4k
The power absorbed by the 1kV resistor is
4.71× 4.71
P1k =
= 22.18 mW
1k
PROBLEM 1.6
Determine the total current in the circuit shown in Fig. 1.41.
Solution Resistances R2, R3 and R4 are in parallel.
equivalent resistance R5 5 R2 || R3 || R4
1
=
1 / R2 +1 / R3 +1 / R4
R5 5 1 V
R1 and R5 are in series,
Fig. 1.41
equivalent resistance RT 5 R1 1 R5 5 5 1 1 5 6 V
And the total current
IT =
Vs 30
= =5 A
RT
6
29
Circuit Elements and Kirchhoff’s Laws
PROBLEM 1.7
Find the current in the 10 V resistance, V1, and source
voltage Vs in the circuit shown in Fig. 1.42.
Solution Assume voltage at the node C 5 V
By applying Kirchhoff’s current law, we get the current
in the 10 V resistance as
I10 5 I5 1 I6
541155A
Fig. 1.42
The voltage across the 6 V resistor is V6 5 24 V
voltage at the node C is VC 5 – 24 V.
The voltage across the branch CD is the same as the voltage at the node C.
Voltage across 10 V only 5 10 3 5 5 50 V
So
VC 5 V10 – V1
–24 5 50 – V1
V1 5 74 V
Now, consider the loop CABD shown in Fig. 1.43.
If we apply Kirchhoff’s voltage law, we get
Fig. 1.43
Vs 5 5 2 30 2 24 5 2 49 V
PROBLEM 1.8
What is the voltage across A and B in the circuit shown in Fig. 1.44?
Fig. 1.44
Solution The above circuit can be redrawn as shown in Fig. 1.45.
Fig. 1.45
30
Circuits and Networks
Assume loop currents I1 and I2 as shown in Fig. 1.45.
6
I1 = = 0.6 A
10
12
I 2 = = 0.86 A
14
VA 5 Voltage drop across the 4 V resistor 5 0.6 3 4 5 2.4 V
VB 5 Voltage drop across the 4 V resistor 5 0.86 3 4 5 3.44 V
The voltage between points A and B is the sum of
voltages as shown in Fig. 1.46.
VAB 5 –2.4 1 12 1 3.44
Fig. 1.46
5 13.04 V
PROBLEM 1.9
Determine the current delivered by the source in the circuit shown in Fig. 1.47.
Fig. 1.47
Solution The circuit can be modified as shown in Fig. 1.48, where R10 is the series combination of R2 and R3.
Fig. 1.48
R10 5 R2 1 R3 5 4 V
R11 is the series combination of R4 and R5.
R11 5 R4 1 R5 5 3 V
Further simplification of the circuit leads to Fig. 1.49 where R12 is the parallel combination of R10 and R9.
R12 5 (R10 || R9) 5 (4 || 4) 5 2 V
31
Circuit Elements and Kirchhoff’s Laws
Fig. 1.49
Fig. 1.50
Similarly, R13 is the parallel combination of R11 and R8.
R13 5 (R11 || R8) 5 (3 || 2) 5 1.2 V
In Fig. 1.49 as shown, R12 and R13 are in series, which is in parallel with R7 forming R14. This is shown in
Fig. 1.50.
R14 5 [(R12 1 R13) || R7]
5 [(2 1 1.2) || 2] 5 1.23 V
Further, the resistances R14 and R6 are in series, which is in parallel with R1 and gives the total resistance
RT 5 [(R14 1 R6) || R1]
5 [(1 1 1.23) || (2)] 5 1.05 V
The current delivered by the source 5 30/1.05 5 28.57 A
PROBLEM 1.10
Determine the current in the 10 V resistance and find Vs in the circuit shown in Fig. 1.51.
Fig. 1.51
Solution The current in the 10 V resistance
I10 5 total current 3 (RT)/(RT 1 R10)
where RT is the total parallel resistance.
7
I10 = 4× =1.65 A
17
Similarly, the current in the resistance R5 is
10
I 5 = 4×
= 2.35 A
10 + 7
or 4 2 1.65 5 2.35 A
The same current flows through the 2 V resistance.
voltage across the 2 V resistance, Vs 5 I5 3 2
5 2.35 3 2 5 4.7 V
32
Circuits and Networks
PROBLEM 1.11
Determine the value of the resistance R and current in each branch when
the total current taken by the circuit shown in Fig. 1.52 is 6 A.
Solution The current in the branch ADB
Fig. 1.52
I30 5 50/(25 1 5) 5 1.66 A
The current in the branch ACB, I10 1 R 5 50/ (10 1 R).
According to Kirchhoff’s current law,
IT 5 I30 1 I10 1 R
6A 5 1.66 A 1 I10 1 R
I10 1 R 5 6 – 1.66 5 4.34 A
∴
50
= 4.34
10 + R
50
10 + R =
=11.52
4.34
R 5 1.52 V
PROBLEM 1.12
Find the power delivered by the source in the circuit shown in Fig. 1.53.
Fig. 1.53
Solution Between points C(E) and D, resistances R3 and R4 are in parallel, which gives
R8 5 (R3 || R4) 5 2.5 V
33
Circuit Elements and Kirchhoff’s Laws
Between points B and C(E), resistances R2 and R7 are in parallel, which gives
R9 5 (R2 || R7) 5 1.5 V
Between points C(E) and D, resistances R6 and R8 are in parallel which gives
R10 5 (R6 || R8) 5 1.25 V
The series combination of R1 and R9 gives
R11 5 R1 1 R9 5 3 1 1.5 5 4.5 V
Similarly, the series combination of R5 and R10 gives
R12 5 R5 1 R10 5 5.25 V
The resistances R11 and R12 are in parallel, which gives
Total resistance 5 (R11 || R12) 5 2.42 ohms
These reductions are shown in Figs 1.54 (a), (b), (c), and (d).
Fig. 1.54
Current delivered by the source =
10
= 4.13 A
2.42
Power delivered by the source 5 VI
5 10 3 4.13 5 41.3 W
PROBLEM 1.13
Determine the voltage drop across the 10 V resistance in the circuit as
shown in Fig. 1.55.
Fig. 1.55
34
Circuits and Networks
Solution The circuit is redrawn as shown in Fig. 1.56.
This is a single-node pair circuit. Assume voltage VA at the node A. By applying Kirchhoff’s current law
at the node A, we have
VA VA VA
+ + = 10 +15
20 10
5
1
1 1
VA  + +  = 25 A
 20 10 5 
A
(0.05 + 0.1+ 0.2)= 25 A
25
= 71.42 V
0
.35
Fig. 1.56
The voltage across 10 V is nothing but the voltage at the node A.
VA =
V10 5 VA 5 71.42 V
PROBLEM 1.14
In the circuit shown in Fig. 1.57, what are the values of R1 and R2, when the current flowing through R1 is 1
A and R2 is 5 A? What is the value of R2 when the current flowing through R1 is zero?
Fig. 1.57
Solution The current in the 5 V resistance
I5 5 I1 1 I2 5 1 1 5
56A
Voltage across the resistance 5 V is V5 5 5 3 6 5 30 V
The voltage at the node A, VA 5 100 – 30 5 70 V
∴
I2 =
VA − 30 70 − 30
=
R2
R2
R2 =
70 − 30 40
= =8 V
I2
5
Similarly, R1 =
70 − 50 20
= = 20 V
1
I1
Circuit Elements and Kirchhoff’s Laws
35
When VA 5 50 V, the current I1 in the resistance R1 becomes zero.
50 − 30
∴ I2 =
R2
where I2 becomes the total current.
100 −VA 100 − 50
∴ I2 =
=
=10 A
5
5
∴
R2 =
20 20
= =2 V
I 2 10
PROBLEM 1.15
Determine the output voltage Vout in the circuit shown in Fig. 1.58.
Fig. 1.58
Solution The circuit shown in Fig. 1.58 can be redrawn as shown in Fig. 1.59.
Fig. 1.59
In Fig. 1.59, R2 and R3 are in parallel, R4 and R5 are in parallel. The complete circuit is a single-node pair
circuit. Assuming voltage VA at the node A and applying Kirchhoff’s current law in the circuit, we have
10 A −
VA
V
− 5A − A = 0
4.43
2.67
 1
1 
= 5A
∴ VA 
+
 4.43 2.67 
36
Circuits and Networks
VA [ 0.225 + 0.375]= 5
∴ VA =
5
= 8.33 V
0.6
PROBLEM 1.16
Determine the voltage VAB in the circuit shown in Fig. 1.60.
Fig. 1.60
Solution The circuit in Fig. 1.60 can be redrawn as shown in Fig. 1.61 (a).
Fig. 1.61 (a)
At the node 3, the series combination of R7 and R8 are in parallel with R6, which gives R9 5 [(R7 1 R8) ||
R6] 5 3 V.
At the node 2, the series combination of R3 and R4 are in parallel with R2, which gives R10 5 [(R3 1 R4)
|| R2] 5 3 V.
It is further reduced and is shown in Fig. 1.61 (b).
Simplifying further, we draw it as shown in Fig. 1.61 (c).
Total current delivered by the source =
=
100
RT
100
= 20.22 A
(13 || 8)
13
Current in the 8 V resistor is I 8 = 20.2×
=12.5 A
13 + 8
37
Circuit Elements and Kirchhoff’s Laws
Fig. 1.61 (b)
Fig. 1.61 (c)
8
Current in the 13 V resistor is I13 = 20.2×
= 7.69 A
13 + 8
So I5 5 12.5 A, and I10 5 7.69 A
Current in the 4 V resistance, I4 5 3.845 A
Current in the 3 V resistance, I3 5 6.25 A
VAB 5 VA – VB
where VA 5 I3 3 3 V 5 6.25 3 3 5 18.75 V
VB 5 I4 3 4 V 5 3.845 3 4 5 15.38 V
VAB 5 18.75 – 15.38 5 3.37 V
PROBLEM 1.17
Determine the value of R in the circuit shown in Fig. 1.62,
when the current is zero in the branch CD.
Fig. 1.62
Solution The current in the branch CD is zero, if the potential difference across CD is zero.
That means, voltage at the point C 5 voltage at the point D.
Since no current is flowing, the branch CD is open-circuited. So the same voltage is applied across ACB
and ADB.
10
V10 =VA ×
15
VR =VA ×
R
20 + R
V10 5 VR
and
10
R
V A × =V A ×
15
20 + R
R 5 40 V
38
Circuits and Networks
PROBLEM 1.18
Find the power absorbed by each element in the circuit shown in Fig. 1.63.
Fig. 1.63
Solution Power absorbed by any element 5 VI
where V is the voltage across the element and I is the current passing through that element.
Here, potential rises are taken as (–) sign.
Power absorbed by the 10 V source 5 –10 3 2 5 –20 W
Power absorbed by the resistor R1 5 24 3 2 5 48 W
Power absorbed by the resistor R2 5 14 3 7 5 98 W
Power absorbed by the resistor R3 5 –7 3 9 5 –63 W
Power absorbed by dependent voltage source 5 (1 3 –7) 3 9 5 –63 W
PROBLEM 1.19
Show that the algebraic sum of the five absorbed power values in Fig. 1.64 is zero.
Fig. 1.64
Solution Power absorbed by the 2 A current source 5 (– 4) 3 2 5 – 8 W
Power absorbed by the 4 V voltage source 5 (– 4) 3 1 5 – 4 W
Power absorbed by the 2 V voltage source 5 (2) 3 3 5 6 W
Power absorbed by the 7 A current source 5 (7) 3 2 5 14 W
Power absorbed by the 2ix dependent current source 5 (– 2) 3 2 3 2 5 – 8 W
Hence, the algebraic sum of the five absorbed power values is zero.
39
Circuit Elements and Kirchhoff’s Laws
PROBLEM 1.20
For the circuit shown in Fig. 1.65, find the power absorbed by each of the elements.
Fig. 1.65
Solution The above circuit can be redrawn as shown in Fig. 1.66.
Fig. 1.66
Assume loop current I as shown in Fig. 1.66.
If we apply Kirchhoff’s voltage law, we get
–12 1 I – 2v1 1 v1 1 4I 5 0
The voltage across the 3 V resistor is v1 5 3 I
Substituting v1 in the loop equation, we get I 5 6 A
Power absorbed by the 12 V source 5 (–12) 3 6 5 –72 W
Power absorbed by the 1 V resistor 5 6 3 6 5 36 W
Power absorbed by 2v1 dependent voltage source 5 (2v1)I 5 2 3 3 3 6 3 6
5 –216 W
Power absorbed by the 3 V resistor 5 v1 3 I 5 18 3 6 5 108 W
Power absorbed by the 4 V resistor 5 4 3 6 3 6 5 144 W
PROBLEM 1.21
For the circuit shown in Fig. 1.67, find the power absorbed
by each element.
Fig. 1.67
40
Circuits and Networks
Solution The circuit shown in Fig. 1.67 is a parallel circuit and consists of a single node A. By assuming
voltage V at the node A, we can find the current in each element.
According to Kirchhoff’s current law,
i3 2 12 2 2i2 2 i2 5 0
By using Ohm’s law, we have
i3 =
V
−V
, i2 =
3
2
1
1
V  +1+  = 12
 3
2 
12
∴ V=
= 6.56
1.83
−6.56
6.56
=− 3.28 A
i3 =
= 2.187 A; i2 =
3
2
Power absorbed by the 3 V resistor 5 (16.56) (2.187) 5 14.35 W
Power absorbed by the 12 A current source 5 (– 6.56) 12 5 – 78.72 W
Power absorbed by the 2i2 dependent current source
5 (– 6.56) 3 2 3 (– 3.28) 5 43.03 W
Power absorbed by the 2 V resistor 5 (– 6.52) (– 3.28) 5 21.51 W
PSpice Problems
PROBLEM 1.1
Determine the total current in the following circuit using
PSpice (Fig. 1.68).
Fig. 1.68
* PROGRAM TO CALCULATE TOTAL CURRENT
VS 1 0 DC 30V
R1 1 2 5OHM
R2 2 0 4OHM
R3 2 0 2OHM
R4 2 0 4OHM
.OP
.END
OUTPUT
**** SMALL SIGNAL BIAS SOLUTION
TEMPERATURE 5 27.000 DEG C
41
Circuit Elements and Kirchhoff’s Laws
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
(1) 30.0000
(2) 5.0000
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
VS
–5.000E 1 00
Result
Total current in the circuit is –5 A from the node 1 to the node 0.
Total current in the circuit is 5 A. ( from the node 0 to the node 1).
.OP statement calculates the dc operating point and displays all the node voltages with respect to ground
node with currents through all the voltage sources in the circuit in the output file. The output can be observed
by opening the output file.
PROBLEM 1.2
Determine current in the 10 V resistance and find Vs in the
following circuit (Fig. 1.69).
Fig. 1.69
* TO DETERMINE THE CURRENT IN A RESISTOR
I1 0 1 DC4A
R1 1 2 2OHM
R2 2 4 10OHM
R3 2 3 5OHM
R4 3 0 2OHM
VX 400 V
.OP
.END
OUTPUT
SMALL SIGNAL BIAS SOLUTION TEMPERATURE 5 27.000 DEG C
****************************************************************
NODE VOLTAGE
NODE VOLTAGE
NODE VOLTAGE
NODE VOLTAGE
(1) 24.4710 (2) 16.4710
(3) 4.7059
(4) 0.0000
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
VX
1.647E 1 00
Result
Current in the 10 V resistance is current through Vx 5 1.647 A from 4 to 0 and
Vs is voltage across the node ‘3’ 5 4.7059 V.
In order to view the current through a resistor, with bias point calculation (.OP), an additional voltage
source of 0 V is inserted in series with the element.
42
Circuits and Networks
PROBLEM 1.3
What is the voltage across A and B in the circuit shown in Fig. 1.70?
Fig. 1.70
* PROGRAM TO CALCULATE VAB
V1 0 1 DC 6V
R1 1 2 6OHM
R2 2 0 4OHM
V2 0 3 DC12V
V3 3 4 12V
R3 3 5 4OHM
R4 4 5 10OHM
.OP
.DC V1 661; DC ANALYSIS FOR V 5 6V
.PRINT DC V(2,5); TO PRINT VAB DIRECTLY FROM DC ANALYSIS
.END
OUTPUT
**** DC TRANSFER CURVES
TEMPERATURE 5 27.000 DEG C
****************************************************************
V1
V(2,5)
6.000E 1 00
1.303E 1 01
**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE 5 27.000 DEG C
****************************************************************
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
(1) – 6.0000 (2) – 2.4000 (3) – 12.0000 (4) – 24.0000 (5) – 15.4290
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
– 6.000E – 01
V2
4.441E – 16
V3
– 8.571E – 01
TOTAL POWER DISSIPATION 1.39E 1 01 WATTS
Result
VAB 5 – V(4V) 1 12 1 V(4V) 5 (– 0.6X4) 1 12 – (4X – 0.857) 5 13.028 V.
PROBLEM 1.4
Determine the output voltage in the following circuit (Fig.1.71).
Circuit Elements and Kirchhoff’s Laws
Fig. 1.71
* PROGRAM TO CALCULATE OUTPUT VOLTAGE
I1 0 1 10A
R1 1 2 3
R2 2 0 2
R3 2 0 5
I2 1 0 5
R4 3 1 10
R5 3 1 2
R6 3 0 1
.OP
.END
**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE 5 27.000 DEG C
****************************************************************
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
(1) 8.3221
(2) 2.6846
(3) 3.1208
VOLTAGE SOURCE CURRENTS
NAME CURRENT
Result
Output voltage Vout 5 8.3221 V 5 V (1).
PROBLEM 1.5
Use PSpice to calculate all the voltages and currents in the following circuit (Fig. 1.72).
Fig. 1.72
* COMPUTATION OF VOLTAGES AND CURRENTS
V1 1 0 DC 60
43
44
Circuits and Networks
R1 1
R2 1
VX 3
G1 0
F1 0
.OP
.END
3
2
0
2
2
20
5
DC 0V
1 3 0.08333
VX 11.6667
OUTPUT
**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE 5 27.000 DEG C
****************************************************************
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
(1) 60.0000 (2) 260.0000 (3) 0.0000
VOLTAGE SOURCE CURRENTS
NAME CURRENT
V1
3.700E 1 01
VX
3.000E 1 00
Result
i1 5 –37 A; i2 5 3 A; i5 5 –35 A; i4 5 –5 A.
V1 5 –60 V; V2 5 60 V; V3 5 –200 V; V4 5 V5 5 260 V.
ANSWERS TO PRACTICE PROBLEMS
1-6.1
1-2.4
(a) 75 A (b) 20 A (c) 2.5 A; 2 S
0.3 3 1022 J
1-2.6
638 mW
1-2.7
(a) P5A 5 21.389 kW, P10 mS 5 771.6 W, P40 mS 5 3.08 kW
Pdependent 5 22.469 kW
(b) P5A 5 2775.9 W, P10 mS 5 240.8 W, P40 mS 5 963.1 W
Pdependent 5 2428.1 W
1-2.8
P0.2 5 2148.8 W, P20 5 21090.9 W, P4 5 743.8 W, P6 5 495.9 V
1-2.9
1-4.1
(a) 0.156 watts (b) 0.14 watts
3.33 V
1-6.2
1.5 mF
1-8.2
V1 5 V2 5 V3 5 100 V
1-8.4
10 V; 30 V
1-8.5
25 V; 5 V
1-9.4
150 V
1-9.5
(a) 60 V; (b) 213.3 V; (c) 51.79 V
1-9.7
0.682 A; 4.092 A
(c) 5 watts
Circuit Elements and Kirchhoff’s Laws
45
Objective-Type Questions
rrr 1.1
rrr 1.2
rrr 1.3
rrr 1.4
rrr 1.5
rrr 1.6
rrr 1.7
rrr 1.8
rrr 1.9
rrr 1.10
rrr 1.11
rrr 1.12
rrr 1.13
rrr 1.14
rrr 1.15
rrr 1.16
rrr 1.17
rrr 1.18
rrr 1.19
rrr 1.20
rrr 1.21
How many coulombs of charge do 50 3 1031 electrons possess?
(a) 80 3 1012 C
(b) 50 3 1031 C
(c) 0.02 3 10231
(d) 1/80 3 1012 C
Determine the voltage of 100 J/25 C.
(a) 100 V
(b) 25 V
(c) 4 V
(d) 0.25 V
What is the voltage of a battery that uses 800 J of energy to move 40 C of charge through a resistor?
(a) 800 V
(b) 40 V
(c) 25 V
(d) 20 V
Determine the current if a 10-coulomb charge passes a point in 0.5 seconds.
(a) 10 A
(b) 20 A
(c) 0.5 A
(d) 2 A
If a resistor has 5.5 V across it and 3 mA flowing through it, what is the power?
(a) 16.5 mW
(b) 15 mW
(c) 1.83 mW
(d) 16.5 W
Identify the passive element among the following:
(a) Voltage source
(b) Current source
(c) Inductor
(d) Transistor
If a resistor is to carry 1 A of current and handle 100 W of power, how many ohms must it be? Assume that
voltage can be adjusted to any required value.
(a) 50 V
(b) 100 V
(c) 1 V
(d) 10 V
A 100 V resistor is connected across the terminals of a 2.5 V battery. What is the power dissipation in the resistor?
(a) 25 W
(b) 100 W
(c) 0.4 W
(d) 6.25 W
Determine the total inductance of a parallel combination of 100 mH, 50 mH, and 10 mH.
(a) 7.69 mH
(b) 160 mH
(c) 60 mH
(d) 110 mH
How much energy is stored by a 100 mH inductance with a current of 1 A?
(a) 100 J
(b) 1 J
(c) 0.05 J
(d) 0.01 J
Five inductors are connected in series. The lowest value is 5 mH. If the value of each inductor is twice that of
the preceding one, and if the inductors are connected in order of ascending values, what is the total inductance?
(a) 155 mH
(b) 155 H
(c) 155 mH
(d) 25 mH
Determine the charge when C 5 0.001 mF and v 5 1 kV.
(a) 0.001 C
(b) 1 mC
(c) 1 C
(d) 0.001 C
If the voltage across a given capacitor is increased, the amount of stored charge
(a) increases
(b) decreases
(c) remains constant
(d) is exactly doubled
1 mF, a 2.2 mF, and a 0.05 mF capacitors are connected in series. The total capacitance is less than
(a) 0.07
(b) 3.25
(c) 0.05
(d) 3.2
How much energy is stored by a 0.05 mF capacitor with a voltage of 100 V?
(a) 0.025 J
(b) 0.05 J
(c) 5 J
(d) 100 J
Which one of the following is an ideal voltage source?
(a) Voltage independent of current
(b) Current independent of voltage
(c) Both (a) and (b)
(d) None of the above
The following voltage drops are measured across each of three resistors in series: 5.2 V, 8.5 V and 12.3 V.
What is the value of the source voltage to which these resistors are connected?
(a) 8.2 V
(b) 12.3 V
(c) 5.2 V
(d) 26 V
A certain series circuit has a 100 V, a 270 V, and a 330 V resistor in series. If the 270 V resistor is removed,
the current
(a) increases
(b) becomes zero
(c) decrease
(d) remain constant
A series circuit consists of a 4.7 kV, 5.6 kV, 9 kV, and 10 kV resistor. Which resistor has the most voltage
across it?
(a) 4.7 kV
(b) 5.6 kV
(c) 9 kV
(d) 10 kV
The total power in a series circuit is 10 W. There are five equal-value resistors in the circuit. How much power
does each resistor dissipate?
(a) 10 W
(b) 5 W
(c) 2 W
(d) 1 W
When a 1.2 kV resistor, 100 V resistor, 1 kV resistor, and a 50 V resistor are in parallel, the total resistance
is less than
(a) 100 V
(b) 50 V
(c) 1 kV
(d) 1.2 kV
46
Circuits and Networks
rrr 1.22 If a 10 V battery is connected across the parallel resistors of 3 V, 5 V, 10 V, and 20 V, how much
rrr 1.23
rrr 1.24
rrr 1.25
rrr 1.26
rrr 1.27
rrr 1.28
rrr 1.29
rrr 1.30
voltage is there across the 5 V resistor?
(a) 10 V
(b) 3 V
(c) 5 V
(d) 20 V
If one of the resistors in a parallel circuit is removed, what happens to the total resistance?
(a) Decreases
(b) Increases
(c) Remains constant
(d) Exactly doubles
The power dissipation in each of three parallel branches is 1 W. What is the total power dissipation of the circuit?
(a) 1 W
(b) 4 W
(c) 3 W
(d) zero
In a four-branch parallel circuit, 10 mA of current flows in each branch. If one of the branch opens, the current
in each of the other branches
(a) increases
(b) decreases
(c) remains unaffected
(d) doubles
Four equal-value resistors are connected in parallel. Five volts are applied across the parallel circuit, and 2.5
mA are measured from the source. What is the value of each resistor?
(a) 4 V
(b) 8 V
(c) 2.5 V
(d) 5 V
Six lightbulbs are connected in parallel across 110 V. Each bulb is related at 75 W. How much current flows
through each bulb?
(a) 0.682 A
(b) 0.7 A
(c) 75 A
(d) 110 A
A 330 V resistor is in series with the parallel combination of four 1 kV resistors. A 100 V source is connected
to the circuit. Which resistor has the most current through it?
(a) 330 V resistor
(b) Parallel combination of three 1 kV resistors
(c) Parallel combination of two 1 kV resistors
(d) 1 kV resistor
The current i4 in the circuit shown in Fig. 1.73 is equal to
(a) 12 A
(b) –12 A
(c) 4 A
(d) none of the above
The voltage V in Fig. 1.74 is equal to
(a) 3 V
(b) –3 V
(c) 5 V
(d) none of the above
Fig. 1.73
Fig. 1.74
rrr 1.31 The voltage V in Fig. 1.75 is always equal to
(a) 9 V
(b) 5 V
(c) 1 V
(d) none of the above
(c) 5 V
(d) none of the above
rrr1.32 The voltage V in Fig. 1.76 is
(a) 10 V
(b) 15 V
Fig. 1.75
Fig. 1.76
For interactive quiz with answers,
scan the QR code given here
OR
visit
http://qrcode.flipick.com/index.php/259
2
LEARNING OBJECTIVES
2.1
INTRODUCTION
A division of mathematics called topology or graph theory deals with graphs of networks and provides
information that helps in the formulation of network equations. In circuit analysis, all the elements in a network
must satisfy Kirchhoff’s laws, besides their own characteristics. Based on these laws, we can form a number
of equations. These equations can be easily written by converting the network into a graph. Certain aspects
of network behaviour are brought into better perspective if a graph of the
network is drawn. If each element or a branch of a network is represented
on a diagram by a line irrespective of the characteristics of the elements, we
get a graph. Hence, network topology is network geometry. A network is an
interconnection of elements in various branches at different nodes as shown
in Fig. 2.1. The corresponding graph is shown in Fig. 2.2 (a).
The graphs shown in Figs 2.2 (b) and (c) are also graphs of the network
in Fig. 2.1.
It is interesting to note that the graphs shown in Fig. 2.2 (a), (b) and (c)
may appear to be different but they are topologically equivalent. A branch
is represented by a line segment connecting a pair of nodes in the graph
of a network. A node is a terminal of a branch, which is represented by a
point. Nodes are the end points of branches. All these graphs have identical
relationships between branches and nodes.
Fig. 2.1
48
Circuits and Networks
(a)
(b)
(c)
Fig. 2.2
The three graphs in Fig. 2.2 have six branches and four nodes. These graphs
are also called undirected. If every branch of a graph has a direction as shown
in Fig. 2.3, then the graph is called a directed graph.
A node and a branch are incident if the node is a terminal of the branch.
Nodes can be incident to one or more elements. The number of branches
incident at a node of a graph indicates the degree of the node. For example, in
Fig. 2.3 the degree of node 1 is three. Similarly, the degree of node 2 is three.
If each element of the connected graph is assigned a direction as shown in Fig.
2.3 it is then said to be oriented. A graph is connected if and only if there is a
Fig. 2.3
path between every pair of nodes. A path is said to exist between any two nodes,
for example 1 and 4 of the graph in Fig. 2.3, if it is possible to reach the node 4 from the node 1 by traversing
along any of the branches of the graph. A graph can be drawn if there
exists a path between any pair of nodes. A loop exists, if there is more
than one path between two nodes.
‰ Planar‰and‰Non-Planar‰Graphs‰ A graph is said to be planar
if it can be drawn on a plane surface such that no two branches cross
each other. On the other hand, in a non-planar graph, there will be
branches which are not in the same plane as others, i.e. a non-planar
graph cannot be drawn on a plane surface without a crossover.
Figures 2.2 and 2.4 illustrate a planar graph and non-planar graph
respectively.
Fig. 2.4
2.2
TREE AND CO-TREE
A tree is a connected subgraph of a network which consists of all the nodes of the
original graph but no closed paths. The graph of a network may have a number
of trees. The number of nodes in a graph is equal to the number nodes in the tree.
The number of branches in a tree is less than the number of branches in a graph. A
graph is a tree if there is a unique path between any pair of nodes. Consider a graph
with four branches and three nodes as shown in Fig. 2.5.
LO 1
49
Methods of Analysing Circuits
Five open-ended graphs based on Fig. 2.5 are represented by Figs 2.6 (a) to (e).
Since each of these open-ended graphs satisfies all the requirements of a tree, each
graph in Fig. 2.6 is a tree corresponding to Fig. 2.5.
In Fig. 2.6, there is no closed path or loop; the number of nodes n 5 3 is the same
for the graph and its tree, where as the number of branches in the tree is only two. In
general, if a tree contains n nodes, then it has (n – 1) branches.
In forming a tree for a given graph, certain branches are removed or opened.
The branches thus opened are called links or link branches. The links for Fig.
2.6 (a) for example are a and d and for Fig. 2.6 (b) are b and c. The set of all links
Fig. 2.5
of a given tree is called the co-tree of the graph. Obviously, the branches a, d are
a co-tree for Fig. 2.6 (a). Similarly, for the tree in Fig. 2.6 (b), the branches b, c are the co-tree. Thus the link
branches and the tree branches combine to form the graph of the entire network.
Fig. 2.6
EXAMPLE 2.1
For the given graph shown in Fig. 2.7, draw the number of possible trees.
Solution‰ The number of possible trees for Fig. 2.7 are represented by Figs 2.8 (a) – (l).
Fig. 2.8
Fig. 2.7
50
2.3
Circuits and Networks
TWIGS AND LINKS
LO 1
The branches of a tree are called its twigs. For a given graph, the complementary set of branches of the
tree is called the co-tree of the graph. The branches of a co-tree are called links, i.e. those elements of the
connected graph that are not included in the tree links and form a subgraph. For example, the set of branches
(b, d, f ) represented by dotted lines in Fig. 2.11 form a co-tree of the graph in Fig. 2.9 with respect to the tree
f
in Fig. 2.10.
a
d
2
1
3
c
e
b
4
Fig. 2.9
Fig. 2.11
Fig. 2.10
The branches a, c, and e are the twigs while the branches b, d, and f are the links of this tree. It can be
seen that for a network with b branches and n nodes, the number of twigs for a selected tree is (n – 1) and
the number of links I with respect to this tree is (b – n 1 1). The number of twigs (n – 1) is known as the
tree value of the graph. It is also called the rank of the tree. If a link is added to the tree, the resulting
graph contains one closed path, called a loop. The addition of each subsequent link forms one or more
additional loops. Loops which contain only one link are independent and are called basic loops.
Frequently Asked Questions linked to LO1*
rrr 2-1.1
(PTU 2011-12)
(PTU 2011-12)
Define tree and co-tree.
rrr 2-1.2 Define loop.
2.4
INCIDENCE MATRIX (A)
The incidence of elements to nodes in a connected graph is shown
LO 2
by the element node incidence matrix (A). Arrows indicated in the
branches of a graph result in an oriented or a directed graph.These arrows
are the indication for the current flow or voltage rise in the network. It
can be easily identified from an oriented graph regarding the incidence of branches to nodes. It is possible to have
an analytical description of an oriented-graph in a matrix form. The dimensions of the matrix A is n 3 b where
n is the number of nodes and b is number of branches. For a graph having n nodes and b branches, the complete
incidence matrix A is a rectangular matrix of order n 3 b.
In the matrix A with n rows and b columns, an entry aij in the i th row and j th column has the following
values.
aij 5 1, if the j th branch is incident to and oriented away from the i th node.
aij 5 –1, if the j th branch is incident to and oriented towards the i th node.
(2.1)
aij 5 0, if the j th branch is not incident to the ith node.
}
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
51
Methods of Analysing Circuits
Figure 2.12 shows a directed graph.
Following the above convention, its incidence matrix A is given by
nodes
↓
a
1
A= 2
3
4
b
branches→
e
c d
f
 1 0
0 1
1 0


0 1
0 0
−1 −1
 0 1
0 0
1 −1 

 0 0 −1 −1 −1 0 


The entries in the first row indicate that three branches a, c, and f are incident to
the node 1 and they are oriented away from the node 1 and therefore the entries a11;
a13 and a16 are 11. Other entries in the first row are zero as they are not connected
to the node 1. Likewise, we can complete the incidence matrix for the remaining
nodes 2, 3, and 4.
2.5
PROPERTIES OF INCIDENCE MATRIX A
Fig. 2.12
LO 2
The following properties are some of the simple conclusions from the incidence matrix A.
1. Each column representing a branch contains two non-zero entries 11 and –1; the rest being zero. The
unit entries in a column identify the nodes of the branch between which it is connected.
2. The unit entries in a row identify the branches incident at a node. Their number is called the degree
of the node.
3. A degree of 1 for a row means that there is one branch incident at the node. This is commonly possible in a tree.
4. If the degree of a node is two, then it indicates that two branches are incident at the node and these
are in series.
5. Columns of A with unit entries in two identical rows correspond to two branches with same end
nodes and hence they are in parallel.
6. Given the incidence matrix A, the corresponding graph can be easily constructed since A is a
complete mathematical replica of the graph.
7. If one row of A is deleted the resulting (n – 1) 3 b matrix is called the reduced incidence matrix
A1. Given A1, A is easily obtained by using the first property.
It is possible to find the exact number of trees that can be generated from a given graph if the reduced
incidence matrix A1 is known and the number of possible trees is given by Det ( A1 A1T ) where A1T is the
transpose of the matrix A1.
EXAMPLE 2.2
Draw the graph corresponding to the given incidence matrix.
 −1 0
0
0 +1 0 +1 0 


 0 −1 0
0
0
0 −1 +1


A =  0
0 −1 −1 0 −1 0 −1
 0
0
0
0 −1 +1 0
0



0
0
0
+1 +1 +1 +1 0
52
Circuits and Networks
Solution‰ There are five rows and eight columns which indicate that there are five nodes and eight branches.
Let us number the columns from a to h and rows as 1 to 5.
a b
c
d
e
f

−
1
0
0
0
1
0
1 
0
0
0
2  0 −1 0

A= 3  0
0 −1 −1 0 −1
 0
0
0
0 −1
1
4 
5  1
1
1
1 0
0
Mark the nodes corresponding to the
rows 1, 2, 3, 4, and 5 as dots as shown
in Fig. 2.13 (a). Examine each column of
A and connect the nodes (unit entries) by
a branch; label it after marking an arrow.
For example, examine the first column
of A. There are two unit entries one
in the first row and 2nd in the last row,
hence connect branch a between node 1
and 5. The entry of A11 is –ve and that
of A51 is 1ve. Hence, the orientation of
the branch is away from the node 5 and
towards node 1 as per the convention.
(a)
Proceeding in this manner, we can
complete the entire graph as shown in Fig.
2.13 (b).
From the incidence matrix A, it can be verified that branches c
branches e and f are in series (property 4).
g
1
−1
h
0

1
0 −1
0
0

0
0
Fig. 2.13
(b)
and d are in parallel (property 5) and
EXAMPLE 2.3
Obtain the incidence matrix A from the following reduced incidence matrix A1 and draw its graph.
−1
1 0
0
0
0

 0 −1
1
1 0
0


[ A1] =  0 0 0 −1 1 0
 0
0
0
0 −1
1


0 −1 0
0 −1
 0
0

0

0
0

1
Solution‰ There are five rows and seven columns in the given reduced incidence matrix [A1]. Therefore, the
number of rows in the complete incidence matrix A will be 5 1 1 5 6. There will be six nodes and seven
branches in the graph. The dimensions of matrix A is 6 3 7. The last row in A, i.e. 6th row for the matrix A
can be obtained by using the first property of the incidence matrix. It is seen that the first column of [A1] has a
single non-zero element –1. Hence, the first element in the 6th row will be 11(–1 1 1 5 0). Second column
of A1 has two non-zero elements 11 and –1, hence the second element in the 6th row will be 0. Proceeding
in this manner we can obtain the 6th row.
53
Methods of Analysing Circuits
The complete incidence matrix can therefore be written as
−1
1 0
0
0
0
0


 0 −1

1
1
0
0
0



c 0
0
0 −1
1 0
0
[ A] = 
d  0
0
0
0 −1
1 0


e 0
0 −1 0
0 −1
1


f  1 0
0
0
0
0 −1
We have seen that any one of the rows of a complete incidence
matrix can be obtained from the remaining rows. Thus, it is possible to
delete any one row from A without loosing any information in A1. Now
the oriented graph can be constructed from the matrix A. The nodes
may be placed arbitrarily. The number of nodes to be marked will be
six. Taking node 6 as reference node the graph is drawn as shown in Fig. 2.14.
a
b
2.6
Fig. 2.14
INCIDENCE MATRIX AND KCL
LO 2
Kirchhoff’s Current Law (KCL) of a graph can be expressed in terms of the reduced incidence matrix as A1
I 5 0.
A1, I is the matrix representation of KCL, where I represents branch current vectors I1, I2, … I6.
Consider the graph shown in Fig. 2.15. It has four nodes a, b, c, and d.
Let node d be taken as the reference node. The positive reference direction of the branch currents
corresponds to the orientation of the graph branches. Let the branch currents be i1, i2, … i6. Applying KCL
at nodes a, b, and c.
– i1 1 i4 5 0
– i2 – i4 1 i5 5 0
– i3 – i5 – i6 5 0
These equations can be written in the matrix form as follows:
−1 0
0
1 0
0


 0 −1 0 −1
1 0

 0
0 −1 0 −1 −1

 i1 
 
i2 
   0
i   
 3  =  0
i   
 4   0
   
i5 
i6 
A1 Ib 5 0
Here, Ib represents column matrix or a vector of branch currents.
 i1 
 
i 
Ib =  2 
�
i 
 b
Fig. 2.15
(2.2)
54
Circuits and Networks
A1 is the reduced incidence matrix of a graph with n nodes and b branches. And it is a (n – 1) 3 b matrix
obtained from the complete incidence matrix of A deleting one of its rows. The node corresponding to the
deleted row is called the reference node or datum node. It is to be noted that A1 Ib 5 0 gives a set of n – 1
linearly independent equations in branch currents i1, i2, … i6. Here, n 5 4. Hence, there are three linearly
independent equations.
Frequently Asked Questions linked to LO2
rrr 2-2.1
Define tree, co-tree, twig, link , and incidence matrix taking a suitable example. (PTU 2009-10)
Define incidence matrix.
(PTU 2009-10)
rrr 2-2.3 Explain the formulation of graph, tree, and incidence matrix using suitable examples. Hence,
discuss the procedure of forming reduced incidence matrix and its advantages. [GTU Dec. 2012]
rrr 2-2.4 Explain about linear oriented graph, incidence matrix and circuit matrix. show Kirchhoff’s laws in
incidence-matrix formulation and circuit-matrix formulation.
[GTU Dec. 2010]
rrr 2-2.2
2.7
LINK CURRENTS: TIE-SET MATRIX
For a given tree of a graph, addition of each link between any two nodes forms
LO 3
a loop called the fundamental loop. In a loop there exists a closed path and a
circulating current, which is called the link current. The current in any branch of
a graph can be found by using link currents.
The fundamental loop formed by one link has a unique path in the tree joining the two nodes of the link.
This loop is also called f-loop or a tie-set.
Consider a connected graph shown in Fig. 2.16 (a). It has four nodes and six branches. One of its trees
is arbitrarily chosen and is shown in Fig. 2.16 (b). The twigs of this tree are branches 4, 5, and 6. The links
corresponding to this tree are branches 1, 2, and 3. Every link defines a fundamental loop of the network.
Number of nodes n 5 4
Number of branches b 5 6
Number of tree branches or twigs 5 n – 1 5 3
Number of link branches I 5 b – (n – 1) 5 3
Let i1, i2, … i6 be the branch currents with directions as shown in Fig. 2.16 (a). Let us add a link in its
proper place to the tree as shown in 2.16(c). It is seen that a loop I1 is formed by the branches 1, 5, and 6.
There is a formation of link current, let this current be I1. This current passes through the branches 1, 5, and
6. By convention a fundamental loop is given the same orientation as its defining link, i.e. the link current
I1 coincides with the branch current direction i1 in ab. A tie set can also be defined as the set of branches
Fig. 2.16
55
Methods of Analysing Circuits
that forms a closed loop in which the link current flows. By adding the other link branches 2 and 3, we can
form two more fundamental loops or f-loops with link currents I2 and I3 respectively as shown in Figs 2.16
(d) and (e).
Fig. 2.16 (continued)
2.7.1 Tie-Set Matrix
Kirchhoff’s voltage law can be applied to the f-loops to get a set of
linearly independent equations. Consider Fig. 2.17.
There are three fundamental loops I1, I2 and I3 corresponding
to the link branches 1, 2, and 3 respectively. If V1, V2, … V6 are
the branch voltages the KVL equations for the three f-loops can be
written as
V1 1 V5 – V6 5 0
V2 1 V4 – V5 5 0
V3 – V4 5 0
}
(2.3)
Fig. 2.17
In order to apply KVL to each fundamental loop, we take the reference direction of the loop which
coincides with the reference direction of the link defining the loop.
The above equation can be written in matrix form as
loop branches →
3×6 ×6
↓ 1 2 3
4
5
6
0
1 −1  V1 
I1  1 0 0


 

1 −1 0  V2 
I 2 0 1 0
0
0  V3   
I 3  0 0 1 −1 0
=
 V   0 
 4   0 


 V5 
V 
 6 
B Vb 5 0
(2.4)
where B is an I 3 b matrix called the tie-set matrix or fundamental loop matrix and Vb is a column vector of
branch voltages.
The tie-set matrix B is written in a compact form as B [bij].
(2.5)
The element bij of B is defined as
bij 5 1 when, the branch bj is in the f-loop Ii (loop current) and their reference directions coincide.
56
Circuits and Networks
bij 5 21 when the branch bj is in the f-loop Ii (loop current) and their reference directions are opposite.
bij 5 0 when branch bj is not in the f-loop Ii.
2.7.2 Tie-set Matrix and Branch Currents
It is possible to express branch currents as a linear combination of link current using the matrix B.
If IB and II represents the branch-current matrix and loop-current matrix respectively and B is the tie-set
matrix, then
[Ib] 5 [BT ] [IL]
(2.6)
where [BT] is the transpose of the matrix [B]. Equation (2.6) is known as link current transformation equation.
Consider the tie-set matrix of Fig. 2.17.
1 0

B =  0 1
0 0

 1

 0

 0
T
B = 
 0

 1

−1
0
0
1
0
1
0
1
−1
0
0
1 −1

1 −1 0
−1 0
0
0

0 
1 
−1 

0

0 
The branch-current vector [Ib] is a column vector.
i1 
 
i2 
 
i
[ I b ]= i3 
 4
i 
 5
i6 
The loop-current vector [IL] is a column vector.
 I1 
 
[ I L ]=  I 2 
 
 I 3 
Therefore, the link-current transformation equation is given by [Ib] 5 [BT ] [IL]
i1   1 0
0 
  

i2   0

1
0
  
 I 
i   0
0
1   1 
 3 = 
 I2 
i   0
1 −1   I 
 4 
  
  3 
i5   1 −1 0 
  

0 
i6  −1 0
57
Methods of Analysing Circuits
The branch currents are
i1 5 I1
i2 5 I2
i3 5 I3
i4 5 I2 – I3
i5 5 I1 – I2
i6 5 – I1
EXAMPLE 2.4
For the electrical network shown in Fig. 2.18 (a), draw its topological graph and write its incidence matrix,
tie-set matrix, link current transformation equation and branch currents.
Fig. 2.18 (a)
Solution‰ The voltage source is short-circuited, the current source is open-
circuited and the points which are electrically at same potential are combined
to form a single node. The graph is shown in Fig. 2.18 (b).
Combining the simple nodes and arbitrarily selecting the branch current
directions, the oriented graph is shown in Fig. 2.18 (c). The simplified graph
consists of three nodes. Let them be x, y, and z and five branches 1, 2, 3, 4, and 5.
The complete incidence matrix is given by
nodes
branches →
1 2
3
4
5
↓
1 0 −1
x  1 0


1 0
1 0
A = y −1
z  0 −1 −1 −1
1
Let us choose the node z as the reference or datum node for writing the
reduced incidence matrix A1 or we can obtain A1 by deleting the last row
elements in A.
nodes
branches →
↓
1 2 3 4
5
A1 =
x
y
 1 0 1 0 −1


−1 1 0 1 0


Fig. 2.18 (b)
Fig. 2.18 (c)
58
Circuits and Networks
For writing the tie-set matrix, consider the tree in the graph in Fig. 2.18 (c).
Number of nodes n 5 3
Number of branches 5 5
Number of tree branches or twigs 5 n – 1 5 2
Number of link branches I 5 b – (n – 1) 5 5 – (3 – 1) 5 3
The tree shown in Fig. 2.18 (d) consists of two branches 4 and 5 shown with solid lines and the link
branches of the tree are 1, 2, and 3 shown with dashed lines. The tie-set matrix or fundamental loop matrix
is given by
loop
branches →
4 5
↓ 1 2 3
1 1
I1  1 0 0



B = I 2  0 1 0 −1 0
I 3  0 0 1 0 1
To obtain the link-current transformation equation and thereby branch
Fig. 2.18 (d)
currents, the transpose of B should be calculated.
1 0 0 


0
1 0 

BT =  0
0 1 
 1 −1 0 




 1 0 1 
The equation [Ib] 5 [BT ] [IL]
 i1   1 0 0 
  

i2   0
1 0   I1 
  
 
 i  = 0
0 1   I 2 
 3 
i   1 −1 0   I 3 
 4 
 
  

i
 5   1 0 1 
The branch currents are given by
i1 5 I1
i2 5 I2
i3 5 I3
i4 5 I1 – I2
5 5 I1 1 I3
Frequently Asked Questions linked to LO3
rrr 2-3.1
Draw a tree of the network in Fig. Q.1 taking the branches
denoted by (b2), (b4), and (b5) as tree branches. Give the
fundamental loop matrix. Determine the matrix loop equation
from the fundamental loop matrix. Branch impedances are in
ohms.
(GTU May 2011)
Fig. Q.1
59
Methods of Analysing Circuits
rrr 2-3.2
Define the following terms:
i. Node ii. Tree iii. Incidence matrix iv. Basic tie-set
rrr 2-3.3 Find the branch currents shown in Fig. Q.3 by using the
concept of the tie-set matrix.
5‡
(JNTU Nov. 2012)
Derive the relationship between fundamental tie-set
matrix, impedance matrix, loop current matrix, and loop
emf matrix.
rrr 2-3.5 For the network shown in Fig. Q.5, write down the tie-set matrix.
Fig. Q.3
(PTU 2009-10)
(PTU 2009-10)
5‡
10
l+
1V
10
‡
1‡
6‡
2‡
rrr 2-3.4
5‡
2‡
5‡
+
25 V
–
‡
‡
10
5‡
Fig. Q.5
rrr 2-3.6
rrr 2-3.7
What is loop matrix?
Write down the fundamental loop matrix of the network shown in Fig. Q.7.
(PTU 2011-12)
(PTU 2011-12)
Fig. Q.7
rrr 2-3.8
Draw the oriented graph of the network shown in Fig. Q.8. Select loop current variables and write
the network- equilibrium equation in matrix form.
(PTU 2011-12)
+
3H
– V
2H
1‡
3H
Fig. Q.8
rrr 2-3.9
For the network shown in Fig. Q.9 draw the graph and write down the tie-set matrix.
(RGTU Dec. 2013)
1‡
2‡
2‡
+
2‡
10 V
–
2‡ 1‡
Fig. Q.9
60
2.8
Circuits and Networks
CUT-SET AND TREE BRANCH VOLTAGES
A cut-set is a minimal set of branches of a connected graph such that the removal
LO 4
of these branches causes the graph to be cut into exactly two parts. The important
property of a cut-set is that by restoring any one of the branches of the cut-set, the
graph should become connected. A cut-set consists of one and only one branch of
the network tree, together with any links which must be cut to divide the network into two parts.
Consider the graph shown in Fig. 2.19 (a).
If the branches 3, 5, and 8 are removed from the graph, we see that the connected graph of Fig. 2.19 (a)
is separated into two distinct parts, each of which is connected as shown in Fig. 2.19 (b). One of the parts is
just an isolated node. Now suppose the removed branch 3 is replaced, all others still removed. Figure 2.19 (c)
shows the resultant graph. The graph is now connected. Likewise, replacing the removed branches 5 and 8 of
the set {3, 5, 8} one at a time, all other ones remaining removed, we obtain the resulting graphs as shown in
Figs 2.19 (d) and (e). The set formed by the branches 3, 5, and 8 is called the cut-set of the connected graph
of Fig. 2.19 (a).
Fig. 2.19
2.8.1 Cut-Set Orientation
A cut-set is oriented by arbitrarily selecting the direction. A cut-set divides
a graph into two parts. In the graph shown in Fig. 2.20, the cut-set is {2, 3}.
It is represented by a dashed line passing through branches 2 and 3. This
cut-set separates the graph into two parts shown as part-1 and part 2.
We may take the orientation either from part-1 to part-2 or from part-2 to
part-1.
The orientation of some branches of the cut-set may coincide with
the orientation of the cut-set while some branches of the cut-set may not
Fig. 2.20
61
Methods of Analysing Circuits
coincide. Suppose we choose the orientation of the cut-set {2, 3} from part-1 to part-2 as indicated in Fig.
2.20, then the orientation of the branch 2 coincides with the cut-set, whereas the orientation of the branch 3
is opposite.
2.8.2 Cut-Set Matrix and KCL for Cut-Sets
KCL is also applicable to a cut-set of a network. For any lumped
electrical network, the algebraic sum of all the cut-set branch currents
is equal to zero. While writing the KCL equation for a cut-set, we
assign positive sign for the current in a branch if its direction coincides
with the orientation of the cut-set and a negative sign to the current
in a branch whose direction is opposite to the orientation of the cutset. Consider the graph shown in Fig. 2.21. It has five branches and
four nodes. The branches have been numbered 1 through 5 and their
orientations are also marked. The following six cut-sets are possible as
shown in Figs 2.22 (a)-(f ).
Fig. 2.21
cut-set C1 : {1, 4}; cut-set C2 : {4, 2, 3}
cut-set C3 : {3, 5}; cut-set C4 : {1, 2, 5}
cut-set C5 : {4, 2, 5}; cut-set C6 : {1, 2, 3}
Fig. 2.22
Applying KCL for each of the cut-set, we obtain the following equations. Let i1, i2 … i6 be the branch
currents.
C1 : i1 − i4 = 0
C2 : − i2 + i3 + i4 = 0
C3 : − i3 + i5 = 0
C4 : i1 − i2 + i5 = 0
C5 : − i2 + i4 +ii5 = 0
C6 : i1 − i2 + i3 = 0



















(2.7)
62
Circuits and Networks
These equations can be put into matrix form as
1 0
0 −1 0 0  i1   0

   
 0 −1
 i2   0
1
1
0
0

   
0
0 −1 0 1 0  i3   0

=
 1 −1 0
0 1 0 i4   0


   
1 1 0  i5   0
 0 −1 0

   
1 0 0 0  i6   0
 1 −1
or
QIb 5 0
(2.8)
where the matrix Q is called the augmented cut-set matrix of the graph or all cut-set matrix of the graph. The
matrix Ib is the branch-current vector.
The all cut-set matrix can be written as Q 5 [qij].
where qij is the element in the ith row and jth column. The order of Q is number of cut-sets 3 number of
branch as in the graph.
qij 5 1, if branch j in the cut-set i and the orientation coincides with each other
qij 5 –1, if branch j is in the cut-set i and the orientation is opposite.
qij 5 0, if branch j is not present in cut-set i.
}
(2.9)
EXAMPLE 2.5
For the network-graph shown in Fig. 2.23 (a) with given orientation, obtain the all cut-set (augmented cut-set)
matrix.
Fig. 2.23 (a)
Solution‰ The graph has four nodes and eight branches. There are in all 12 possible cut-sets as shown with
dashed lines in Figs 2.23 (b) and (c). The orientation of the cut-sets has been marked arbitrarily. The cut-sets
are
C1 : {1, 46}; C2 {1, 2, 3}; C3: {2, 5, 8}
C4 : {6, 7, 8}; C5 {1, 3, 5, 8}; C6: {1, 4, 7, 8}
C7 : {2, 5, 6, 7}; C8 : {2, 3, 4, 6} C9 : {1, 4, 7, 5, 2}
C10 : {2, 3, 4, 7, 8} ; C11 : {6, 4, 3, 5, 8}; C12 : {1, 3, 5, 7, 6}
Eight cut-sets C1 to C8 are shown if Fig. 2.23 (b) and four cut-sets C9 to C11 are shown in Fig. 2.23 (c) for
clarity.
63
Methods of Analysing Circuits
Fig. 2.23 (b)
Fig. 2.23 (c)
As explained in Section 2.8.2 with the help of Eq. (2.9), the all cut-set matrix Q is given by
cut-sets
branches →
↓
1
2
3
4 5
6
7
8
C1 −1 0
0
1 0 −1 0
0



C 2  1 −1 −1 0
0
0
0
0



C3
1 0
0
1 0
0 −1
 0
C4  0
0
0
0
0
1
1
1


C5  1 0 −1 0
1 0
0 −1


0
1 0
0
1
1
Q = C6 −1 0


1 0
0
1
1
1 0
C7  0


1 0 −1 0
0
C8  0 −1 −1


C9  1 −1 0 −1 −1 0 −1 0
1
1 −1 0
0 −1 −1
C10  0

0
1 −1 −1
1 0
1
C11  0


1 0 −1 −1 −1 0
C12 −1 0
Matrix Q is a 12 3 8 matrix since there are 12 cut-sets and eight branches in the graph.
2.8.3 Fundamental Cut-Sets
Observe the set of Eq. (2.7) in Section 2.8.2 with respect to the graph in Fig. 2.22. Only the first three equations
are linearly independent, remaining equations can be obtained as a linear combination of the first three. The
concept of fundamental cut-set ( f-cut-set) can be used to obtain a set of linearly independent equations in
branch current variables. The f-cut-sets are defined for a given tree of the graph. From a connected graph,
first a tree is selected, and then a twig is selected. Removing this twig from the tree separates the tree into
two parts. All the links which go from one part of the disconnected tree to the other, together with the twig of
the selected tree, will constitute a cut-set. This cut-set is called a fundamental cut-set or f-cut-set of the graph.
Thus, a fundamental cut-set of a graph with respect to a tree is a cut-set that is formed by one twig and a
unique set of links. For each branch of the tree, i.e. for each twig, there will be a f-cut-set. So, for a connected
graph having n nodes, there will be (n – 1) twigs in a tree, the number of f-cut-sets is also equal to (n – 1).
64
Circuits and Networks
The fundamental cut-set matrix Qf is one in which each row represents a cut-set with respect to a
given tree of the graph. The rows of Q1 correspond to the fundamental cut-sets and the columns correspond
to the branches of the graph. The procedure for obtaining a fundamental cut-set matrix is illustrated in
Example 2.6.
EXAMPLE 2.6
Obtain the fundamental cut-set matrix Qf for the network graph shown in Fig. 2.23 (a).
Solution‰ A selected tree of the graph is shown in Fig. 2.24 (a).
(a)
(b)
Fig. 2.24
The twigs of the tree are {3, 4, 5, 7}. The remaining branches 1, 2, 6, and 8 are the links, corresponding to the
selected tree. Let us consider the twig 3. The minimum number of links that must be added to twig 3 to form a
cut-set C1 is {1, 2}. This set is unique for C1. Thus, corresponding to the twig 3. The f-cut-set C1 is {1, 2, 3}. This
is shown in Fig. 2.24 (b). As a convention the orientation of a cut-set is chosen to coincide with that of its defining
twig. Similarly, corresponding to the twig 4, the f-cut-set C2 is {1, 4, 6} corresponding to the twig 5, the f-cut-set C3
is {2, 5, 8} and corresponding to the twig 7, the f-cut-set is {6, 7, 8}. Thus, the f-cut-set matrix is given by
f -cut-sets
↓
C1  −1

C 2  −1

Qf =
C3  0
C4  0
branches→
1 1 0
0 0 1
0

0
0 0 −1
1 1
1
0
0 0
0 −1 0
1 0 0 +1
0 0 0
0
(2.10)
2.8.4 Tree Branch Voltages and f-Cut-Set Matrix
From the cut-set matrix, the branch voltages can be expressed in terms of tree branch voltages. Since all tree
branches are connected to all the nodes in the graph, it is possible to trace a path from one node to any other
node by traversing through the tree-branches.
Let us consider Example 2.6, there are eight branches. Let the branch voltages be V1, V2, … V8. There are
four twigs, let the twig voltages be Vt3, Vt4, Vt5 and Vt7 for twigs 3, 4, 5 and 7 respectively.
We can express each branch voltage in terms of twig voltages as follows.
V1 5 – V3 – V4 5 – Vt3 – Vt4
65
Methods of Analysing Circuits
V2 5 1 V3 1 V5 5 1 Vt3 1 Vt5
V3 5 Vt3
V4 5 Vt4
V5 5 Vt5
V6 5 V7 – V4 5 Vt7 – Vt4
V7 5 Vt7
V8 5 V7 – V5 5 Vt7 – Vt5
The above equations can be written in matrix form as
 V1  −1 −1
0 0 
  

 V2  +1

   1 0 +1 0 
V   1 0
0 0  Vt3 
 3 
 
V   0
1
0 0  Vt4 
 4 =
(2.11)
  

0
1 0  Vt5 
 V5   0
  

0 1  Vt7 
 V6   0 −1
  

 V7   0
0
0 1 
  

 V8   0
0 −1 1 
The first matrix on the right-hand side of Eq. (2.11) is the transpose of the f-cut-set matrix Qf given in Eq.
(2.10) in Ex. 2.6. Hence, Eq. (2.11) can be written as
Vb = QTf Vt
(2.12)
where Vb is the column matrix of branch-voltages Vt is the column matrix of twig voltages corresponding to
the selected tree and QTf in the transpose of f-cut-set matrix.
Equation (2.12) shows that each branch voltage can be expressed as a linear combination of the treebranch voltages. For this purpose, the fundamental cut-set ( f-cut-set) matrix can be used without writing
loop equations.
Frequently Asked Questions linked to LO4
rrr 2-4.1
For the network shown in Fig. Q.1 draw the oriented graph and all
possible trees. Also prepare (a) incidence matrix, (b) Fundamental
tie-set matrix, (c) Fundamental cut-set matrix. (GTU Dec. 2012)
rrr 2-4.2 Explain the fundamental cut-set matrix taking a suitable example.
(PTU 2009-10)
rrr 2-4.3 Write relation between branch voltage matrix [VO], twig voltage
matrix [VT] and node voltage matrix [VN] in graph theory.
(PTU 2011-12)
Fig. Q.1
rrr
2.9
MESH ANALYSIS
Mesh and nodal analysis are two basic important techniques used in finding
solutions for a network. The suitability of either mesh or nodal analysis to
a particular problem depends mainly on the number of voltage sources or
current sources. If a network has a large number of voltage sources, it is useful
to use mesh analysis; as this analysis requires that all the sources in a circuit
LO 5
66
Circuits and Networks
(a)
(b)
(c)
Fig. 2.25
be voltage sources. Therefore, if there are any current sources in a circuit they are to be converted into equivalent
voltage sources, if, on the other hand, the network has more current sources, nodal analysis is more useful.
Mesh analysis is applicable only for planar networks. For non-planar circuits, mesh analysis is not
applicable. A circuit is said to be planar if it can be drawn on a plane surface without crossovers. A non-planar
circuit cannot be drawn on a plane surface without a crossover.
Figure 2.25 (a) is a planar circuit. Figure 2.25 (b) is a non-planar circuit and Fig. 2.25 (c) is a planar
circuit which looks like a non-planar circuit. It has already been discussed that a loop is a closed path. A
mesh is defined as a loop which does not contain any other loops within it. To apply mesh analysis, our first
step is to check whether the circuit is planar or not and the second is to select mesh currents. Finally, writing
Kirchhoff’s voltage law equations in terms of unknowns and solving them leads to the final solution.
Observation of Fig. 2.26 indicates that there are two loops abefa,
and bcdeb in the network. Let us assume loop currents I1 and I2 with
directions as indicated in the figure. Considering the loop abefa alone,
we observe that current I1 is passing through R1, and (I1 – I2) is passing
through R2. By applying Kirchhoff’s voltage law, we can write
Vs 5 I1R1 1 R2(I1 – I2)
Fig. 2.26
Similarly, if we consider the second mesh bcdeb, the current I2
is passing through R3 and R4, and (I2 – I1) is passing through R2. By
applying Kirchhoff’s voltage law around the second mesh, we have
R2 (I2 – I1) 1 R3 I2 1 R4 I2 5 0
By rearranging the above equations, the corresponding mesh current equations are
I1(R1 1 R2) – I2 R2 5 Vs
– I1 R2 1 (R2 1 R3 1 R4)I2 5 0
By solving the above equations, we can find the currents I1 and I2. If we observe Fig. 2.26, the circuit
consists of five branches and four nodes, including the reference node. The number of mesh currents is equal
to the number of mesh equations.
And the number of equations 5 branches – (nodes – 1). In Fig. 2.26, the required number of mesh currents
would be 5 – (4 – 1) 5 2.
In general, if we have B branches and N nodes including the reference node then the number of linearly
independent mesh equations M 5 B – (N – 1).
67
Methods of Analysing Circuits
EXAMPLE 2.7
Write the mesh current equations in the circuit shown in Fig. 2.27,
and determine the currents.
Solution‰ Assume two mesh currents in the direction as indicated
in Fig. 2.28.
The mesh current equations are
5I1 1 2(I1 – I2) 5 10
Fig. 2.27
10I2 1 2(I2 – I1) 1 50 5 0
We can rearrange the above equations as
7I1 – 2I2 5 10
– 2I1 1 12I2 5 – 50
By solving the above equations, we have
I1 5 0.25 A, and I2 5 – 4.125 A
Here, the current in the second mesh, I2, is negative; that is the
actual current I2 flows opposite to the assumed direction of current
in the circuit of Fig. 2.28.
EXAMPLE 2.8
Determine the mesh current I1 in the circuit shown in Fig. 2.29.
Fig. 2.29
Solution‰ From the circuit, we can form the following three mesh equations:
10I1 1 5(I1 1 I2) 1 3(I1 – I3) 5 50
2I2 1 5(I2 1 I1) 1 1(I2 1 I3) 5 10
3(I3 – I1) 1 1(I3 1 I2) 5 – 5
Rearranging the above equations, we get
18I1 1 5I2 – 3I3 5 50
5I1 1 8I2 1 I3 5 10
– 3I1 1 I2 1 4I3 5 –5
Fig. 2.28
68
Circuits and Networks
According to Cramer’s rule,
 50

 10

 −5
I1 = 
 18
 5

 −3

5 −3 

8
1
1
4  1175
=
5 −3  356
8
1
1
4 
or I1 5 3.3 A
Similarly,
 18 50 −3 


 5 10

1


 −3 −5
4  −355

I2 = 
 = 356
 18 5 −3 
 5 8
1 


4 
 −3 1
or I2 5 – 0.997 A
 18 5 50 


 5 8 10 


 −3 1 −5  525
=
I 3 = 

 18 5 −3  356
 5 8
1

 −3 1
4 

or I3 5 1.47 A
∴ I1 5 3.3 A, I2 5 – 0.997 A, I3 5 1.47 A
2.10
MESH EQUATIONS BY INSPECTION METHOD
LO 5
The mesh equations for a general planar network can be written by inspection without going through the
detailed steps. Consider three mesh networks as shown in Fig. 2.30.
Fig. 2.30
Methods of Analysing Circuits
The loop equations are
I1R1 1 R2(I1 – I2) 5 V1
R2(I2 – I1) 1 I2R3 5 –V2
69
(2.13)
(2.14)
R4I3 1 R5I3 5 V2
(2.15)
Reordering the above equations, we have
(R1 1 R2)I1 – R2I2 5 V1
(2.16)
– R2I1 1 (R2 1 R3)I2 5 –V2
(2.17)
(R4 1 R5)I3 5 V2
(2.18)
The general mesh equations for the three-mesh resistive network can be written as
R11I1  R12I2  R13I3 5 Va
(2.19)
 R21I1 1 R22I2  R23I3 5 Vb
(2.20)
 R31I1  R32I2 1 R33I3 5 Vc
(2.21)
By comparing equations (2.16), (2.17) and (2.18) with Eqs (2.19), (2.20), and (2.21) respectively, the
following observations can be taken into account.
1. The self-resistance in each mesh,
2. The mutual resistances between all pairs of meshes, and
3. The algebraic sum of the voltages in each mesh.
The self-resistance of the loop 1, R11 5 R1 1 R2, is the sum of the resistances through which I1 passes.
The mutual resistance of the loop 1, R12 5 – R2, is the sum of the resistances common to loop currents
I1 and I2. If the directions of the currents passing through the common resistance are the same, the mutual
resistance will have a positive sign; and if the directions of the currents passing through the common resistance
are opposite then the mutual resistance will have a negative sign.
Va 5 V1 is the voltage which drives loop one. Here, the positive sign is used if the direction of the current
is the same as the direction of the source. If the current direction is opposite to the direction of the source,
then the negative sign is used.
Similarly, R22 5 (R2 1 R3) and R33 5 R4 1 R5 are the self resistances of loops two and three, respectively.
The mutual resistances R13 5 0, R21 5 – R2, R23 5 0, R31 5 0, R32 5 0 are the sums of the resistances
common to the mesh currents indicated in their subscripts.
Vb 5 – V2, Vc 5 V2 are the sum of the voltages driving their respective loops.
EXAMPLE 2.9
Write the mesh equations for the circuit shown in Fig. 2.31.
Fig. 2.31
70
Circuits and Networks
Solution‰ The general equations for the three-mesh network are
R11I1  R12I2  R13I3 5 Va
 R21I1 1 R22I2  R23I3 5 Vb
 R31I1  R32I2 1 R33I3 5 Vc
(2.22)
(2.23)
(2.24)
Consider Eq. (2.22).
R11 5 self-resistance of the loop 1 5 (1 V 1 3 V 1 6 V) 5 10 V
R12 5 the mutual resistance common to loops 1 and 2 5 –3 V
Here, the negative sign indicates that the currents are in opposite direction
R13 5 the mutual resistance common to loop 1 and 3 5 –6 V
Va 5 110 V, the voltage driving the loop 1.
Here, the positive sign indicates the loop current I1 is in the same direction as the source element.
Therefore, Eq. (2.22) can be written as
10I1 – 3I2 – 6I3 5 10 V
(2.25)
Consider Eq. (2.23).
R21 5 mutual resistance common to loops 1 and 2 5 –3 V
R22 5 self-resistance of loop 2 5 (3 V 1 2 V 1 5 V) 5 10 V
R23 5 0, there is no common resistance between loops 2 and 3.
Vb 5 –5 V, the voltage driving the loop 2.
Therefore, Eq. (2.23) can be written as
– 3I1 1 10I2 5 – 5 V
(2.26)
Consider Eq. (2.24).
R31 5 mutual resistance common to loops 3 and 1 5 –6 V
R32 5 mutual resistance common to loops 3 and 2 5 0
R33 5 self-resistance of the loop 3 5 (6 V 1 4 V) 5 10 V
Vc 5 algebraic sum of the voltages driving the loop 3
5 (5 V 1 20 V) 5 25 V
Therefore, Eq. (2.24) can be written as
– 6I1 1 10I3 5 25 V
(2.27)
The three mesh equations are
10I1 – 3I2 – 6I3 5 10 V
– 3I1 1 10I2 5 – 5 V
– 6I1 1 10I3 5 25 V
2.11
SUPERMESH ANALYSIS
LO 5
Suppose any of the branches in the network has a current source; then it is slightly difficult to apply mesh
analysis straightforward because first we should assume an unknown voltage across the current source,
writing mesh equations as before, and then relate the source current to the assigned mesh currents. This is
generally a difficult approach. One way to overcome this difficulty is by applying the supermesh technique.
Here we have to choose the kind of supermesh. A supermesh is constituted by two adjacent loops that have
Methods of Analysing Circuits
71
a common current source. As an example, consider the network shown in Fig. 2.32.
Fig. 2.32
Here, the current source I is in the common boundary for the two meshes 1 and 2. This current source
creates a supermesh, which is nothing but a combination of meshes 1 and 2.
R1I1 1 R3(I2 – I3) 5 V
or
R1I1 1 R3I2 – R4I3 5 V
Considering the mesh 3, we have
R3(I3 – I2) 1 R4I3 5 0
Finally, the current I from the current source is equal to the difference between the two mesh currents, i.e.
I1 – I2 5 I
We have, thus, formed three mesh equations which we can solve for the three unknown currents in the
network.
EXAMPLE 2.10
Determine the current in the 5 V resistor in the network given in Fig. 2.33.
Fig. 2.33
Solution‰ From the first mesh, i.e. abcda, we have
50 5 10(I1 – I2) 1 5(I1 – I3)
or 15I1 – 10I2 – 5I3 5 50
From the second and third meshes, we can form a supermesh
10(I2 – I1) 1 2I2 1 I3 1 5(I3 – I1) 5 0
or
– 15I1 1 12I2 1 6I3 5 0
(2.28)
(2.29)
72
Circuits and Networks
The current source is equal to the difference between II and III mesh currents,
i.e. I2 – I3 5 2 A
(2.30)
Solving 2.28, 2.29, and 2.30, we have
I1 5 19.99 A, I2 5 17.33 A, and I3 5 15.33 A
The current in the 5 V resistor 5 I1 – I3
5 19.99 – 15.33 5 4.66 A
∴ the current in the 5 V resistor is 4.66 A.
EXAMPLE 2.11
Write the mesh equations for the circuit shown in Fig. 2.34 and determine the currents, I1, I2 and I3.
Fig. 2.34
Solution‰ In Fig. 2.34, the current source lies on the perimeter of the circuit, and the first mesh is ignored.
Kirchhoff’s voltage law is applied only for second and third meshes.
From the second mesh, we have
3(I2 – I1) 1 2(I2 – I3) 1 10 5 0
or
– 3I1 1 5I2 – 2I3 5 –10
(2.31)
From the third mesh, we have
I3 1 2(I3 – I2) 5 10
or
– 2I2 1 3I3 5 10
(2.32)
From the first mesh,
I1 5 10 A
(2.33)
From the above three equations, we get
I1 5 10 A, I2 5 7.27 A, I3 5 8.18 A
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO5
rr 2-5.1
In the circuit shown in Fig. Q.1, use mesh analysis to find out the power delivered to the 4 V
resistor. To what voltage should the 100 V battery be changed so that no power is delivered to the
4 V resistor?
Methods of Analysing Circuits
73
Fig. Q.1
rrr 2-5.2
Find the voltage between A and B of the circuit shown in Fig. Q.2 by mesh analysis.
Fig. Q.2
rrr 2-5.3
Find the value of R1 and R2 in the network shown in Fig. Q.3 using mesh analysis.
Fig. Q.3
rrr 2-5.4
Using mesh analysis, determine the voltage across the 10 kV resistor at terminals A and B of the
circuit shown in Fig. Q.4.
Fig. Q.4
74
Circuits and Networks
rrr 2-5.5 In
the network shown in Q.5, the resistance R is variable from zero to infinity. The current I
through R can be expressed as I 5 a 1 bV, where V is the voltage across R as shown in the figure,
and a and b are constants. Determine a and b.
Fig. Q.5
rrr2-5.6
For the circuit shown in Fig. Q.6, use mesh analysis to find the values of all mesh currents.
Fig. Q.6
rrr 2.5.7
Determine the value of current I in the following circuit shown in Fig. Q.7 by using mesh analysis.
rrr 2-5.8
Use mesh analysis to find the current supplied by the 31 V source and the current in 4 Ω resistor
of the circuit shown in Fig. Q.8.
Fig. Q.7
Fig. Q.8
75
Methods of Analysing Circuits
rrr2-5.9
For the circuit shown in Fig. Q.9, find the value of V2 that will cause the voltage across 20 V to be
zero by using mesh analysis.
Fig. Q.Q.9
rrr2-5.10
Find the mesh currents I1, I2, I3, and I4 in the following circuit Fig. Q.10, using PSpice
Fig. Q.10
Frequently Asked Questions linked to LO5
rrr 2-5.1
rrr 2-5.2
Distinguish between mesh and loop of an electric circuit.
(AU May/June 2013)
Using mesh analysis, determine the current through the 1 W resistor in the circuit shown in Fig. Q.2
(AU May/June 2013)
Fig. Q.2
rrr 2-5.3
Find out the current in each branch of the circuit shown in Fig. Q.3.
1W
3W
+
5A
5W
10 W
Fig. Q.3
10 V
–
76
rrr 2.5.4
Circuits and Networks
Determine current in each mesh of the circuit shown in Fig. Q.4.
+
10 V
–
3W
10 A
1W
2W
Fig. Q.4
rrr 2-5.5
(AU Nov./Dec. 2012)
(AU Nov./Dec. 2012)
Define mesh analysis of a circuit.
rrr 2-5.6 Determine the current l2 in the circuit shown in Fig. Q.6
3W
+
4V –
3W
I2
3W
–+
8V
1W
5W
+–
1W
6V
Fig. Q.6
rrr2-5.7
Using mesh method obtain the loop currents in the network of Fig. Q.7. What is the total power
loss?
(BPUT 2007)
Fig. Q.7
rrr 2-5.8
Derive a tree of the graph of the network in Fig. Q.8. Determine the node voltage V1 and V2 using
the mesh analysis. Resistance values are in ohms.
(GTU May 2011)
1
2 V dc
2
V1
1 A dc
+
–
2
V2
2
+
4 V dc –
Fig. Q.8
rrr 2-5.9
Find mesh current and determine the voltage across each element in the circuit shown in Fig. Q.9.
(JNTU Nov. 2012)
77
Methods of Analysing Circuits
5W
8W
10 V
5A
4W
2W
Fig. Q.9
rrr 2-5.10 Find the power delivered by the two sources to the circuit shown in Fig. Q.10.
1W
(PTU 2009-10)
2W
+
–
5V
6A
4W
3W
Fig. Q.10
rrr 2-5.11 Determine
method.
the current supplied by each battery in the circuit of Fig. Q.11 using mesh analysis
(PTU 2011-12)
Fig. Q.11
rrr2-5.12
Find the value of K in the circuit shown in Fig. Q.12 such that the power dissipated in the 2 W
resistor does not exceed 50 watts.
(PU 2012)
4W –
KI
+
I
2W
6A
8W
+ 16 V
–
Fig. Q.12
rrr 2-5.13 Determine the current in the 5 W resistor in the network shown in Fig. Q.13.
10 W
2A
+
2W
20 V
–
5W
Fig. Q.13
1W
78
Circuits and Networks
rrr 2-5.14 In
the circuit shown in Fig. Q.14 use loop analysis to find the
power delivered to the 4 W resistor.
2.12
Fig. Q.14
NODAL ANALYSIS
In Chapter 1, we discussed simple circuits containing only two nodes,
LO 6
including the reference node. In general, in an N-node circuit, one of the
nodes is choosen as reference or datum node, then it is possible to write N –
1 nodal equations by assuming N – 1 node voltages. For example, a 10-node
circuit requires nine unknown voltages and nine equations. Each node in a
circuit can be assigned a number or a letter. The node voltage is the voltage
of a given node with respect to one particular node, called the reference
node, which we assume at zero potential. In the circuit shown in Fig. 2.35, the node 3 is assumed as the
reference node. The voltage at the node 1 is the voltage at that node with respect to the node 3. Similarly, the
voltage at the node 2 is the voltage at that node with respect to the node 3. Applying Kirchhoff’s current law
at the node 1; the current entering is equal to the current leaving (see Fig. 2.36).
Fig. 2.36
Fig. 2.35
I1 =
V1 V1 −V2
+
R1
R2
Here, V1 and V2 are the voltages at nodes 1 and 2, respectively. Similarly, at the node 2, the current entering
is equal to the current leaving as shown in Fig. 2.37.
V2 −V1 V2
V2
+ +
=0
R2
R3 R4 + R5
Rearranging the above equations, we have
1
1
1
V1  +  −V2   = I1
R R 
R 
2
 1
 2
Fig. 2.37
1
1
1
1 
−V1   +V2  + +
=0


 
 R2 
 R2 R3 R4 + R5 
Methods of Analysing Circuits
79
From the above equations, we can find the voltages at each node.
EXAMPLE 2.12
Write the node voltage equations and determine the currents in each branch for the network shown in Fig.
2.38.
Fig. 2.38
Solution‰ The first step is to assign voltages at each node as shown in Fig. 2.39.
Fig. 2.39
Applying Kirchhoff’s current law at the node 1, we have
V V −V 2
5= 1 + 1
10
3
 1 1
1
or V 1  +  −V 2   = 5
10 3 
 3 
(2.34)
Applying Kirchhoff’s current law at the node 2, we have
V2 −V1 V2 V2 −10
+ +
=0
3
5
1
or
1
1 1 
−V1   +V2  + +1 =10
 3 
 3 5 
From Eqs (2.34) and (2.35), we can solve for V1 and V2 to get
V1 5 19.85 V, V2 5 10.9 V
I10 =
I5 =
V1
V −V 19.85 −10.9
=1.985 A, I 3 = 1 2 =
= 2.98 A
10
3
3
V2 10.9
V −10
=
= 2.18 A, I1 = 2
= 0.9 A
1
5
5
(2.35)
80
Circuits and Networks
EXAMPLE 2.13
Determine the voltages at each node for the circuit shown in Fig. 2.40.
Fig. 2.40
Solution‰ At the node 1, assuming that all currents are leaving, we have
V1 −10 V1 −V2 V1 V1 −V2
+
+ +
=0
10
3
5
3
1 1
 1 1 1 1
or V1  + + +  −V2  +  =1
10 3 5 3 
 3 3 
0.96V1 – 0.66V2 5 1
(2.36)
At the node 2, assuming that all currents are leaving except the current from the current source, we have
V2 −V1 V2 −V1 V2 −V3
+
+
=5
3
3
2
 2
1 1 1
1
−V1   +V2  + +  −V3   = 5
 3 
 3 3 2 
 2 
– 0.66 V1 1 1.16 V2 – 0.5V3 5 5
(2.37)
At node 3, assuming all currents are leaving, we have
V3 −V2 V3 V3
+ + =0
2
1
6
– 0.5 V2 1 1.66 V3 5 0
Applying Cramer’s rule, we get
 1 −0.66

0


 5
1.16 −0.5 

 0 −0.5
1.66  7.154
V1 = 
=
= 8.06 V
0  0.887
 0.96 −0.66
 −0.66
1.16 −0.5 

 0
−0.5
1.6
66 

(2.38)
Methods of Analysing Circuits
81
Similarly,


0.96 1
0


 −0.66 5 −0.5 




0
0
1
.
66
 = 9.06 = 10.2 V
V2 = 
 0.887
0
.
96
−
0
.
66
0


 −0.66

1
.
16
−
0
.
5


 0
−0.5
1.66 


0.96 −0.66 1 


 −0.66

1
.
16
5



2.73
0
−0.5 0 

=
= 3.07 V
V3 = 

0  0.887
 0.96 −0.66
 −0.66
1.16 −0.5 

 0
−0.5
1.66 

2.13
NODAL EQUATIONS BY INSPECTION METHOD
LO 6
The nodal equations for a general planar network can also be written by inspection, without going through the
detailed steps. Consider a three-node resistive network, including the reference node, as shown in Fig. 2.41.
In Fig. 2.41, the points a and b are the actual nodes and c is the reference node.
Fig. 2.41
Now consider the nodes a and b separately as shown in Fig. 2.42 (a) and (b).
Fig. 2.42
In Fig. 2.42 (a), according to Kirchhoff’s current law, we have
I1 1 I2 1 I3 5 0
82
Circuits and Networks
∴
Va −V1 Va Va −Vb
+ +
=0
R1
R2
R3
(2.39)
In Fig. 2.42 (b), if we apply Kirchhoff’s current law, we get
I4 1 I5 5 I3
∴ Vb −Va + Vb + Vb −V2 = 0
R3
R4
R5
(2.40)
Rearranging the above equations, we get
1

 
 
 + 1 + 1 V − 1 V =  1 V
 a   b   1
 R R
R3 
 R1 
 R3 
1
2
 1
−
 R

1
V
1
1
Va +  + + Vb = 2


 R3 R4 R5 
R5
3
(2.41)
(2.42)
In general, the above equations can be written as
Gaa Va 1 Gab Vb 5 I1
(2.43)
Gba Va 1 Gbb Vb 5 I2
(2.44)
By comparing Eqs (2.41), (2.42), and Eqs (2.43), (2.44) we have the self-conductance at node a, Gaa 5 (1/
R1 1 1/R2 1 1/R3) is the sum of the conductances connected to node a. Similarly, Gbb 5 (1/R3 1 1/R4 1 1/R5), is
the sum of the conductances connected to node b. Gab 5 (–1/R3), is the sum of the mutual conductances connected
to node a and node b. Here all the mutual conductances have negative signs. Similarly, Gba 5 (–1/R3) is also a
mutual conductance connected between nodes b and a. I1 and I2 are the sum of the source currents at the node a
and the node b, respectively. The current which drives into the node has positive sign, while the current that drives
away from the node has negative sign.
EXAMPLE 2.14
For the circuit shown in Fig. 2.43, write the node equations by the inspection method.
Fig. 2.43
Solution‰ The general equations are
Gaa Va 1 Gab Vb 5 I1
(2.45)
Gba Va 1 Gbb Vb 5 I2
(2.46)
Methods of Analysing Circuits
83
Consider Eq. (2.45).
Gaa 5 (1 1 1/2 1 1/3) mho, the self-conductance at the node a is the sum of the conductances connected
to the node a.
Gbb 5 (1/6 1 1/5 1 1/3) mho the self-conductance at the node b is the sum of the conductances connected
to the node b.
Gab 5 – (1/3) mho, the mutual conductance between nodes a and b is the sum of the conductances
connected between nodes a and b.
Similarly, Gba 5 – (1/3), the sum of the mutual conductances between nodes b and a.
I1 =
10
=10 A, the source current at the node a,
1
 2 5
I 2 =  +  =1.23 A, the source current at the node b.
 5 6 
Therefore, the nodal equations are
2.14
1.83 Va – 0.33 Vb 5 10
(2.47)
– 0.33 Va 1 0.7 Vb 5 1.23
(2.48)
SUPERNODE ANALYSIS
LO 6
Suppose any of the branches in the network has a voltage source; then it is slightly difficult to apply nodal
analysis. One way to overcome this difficulty is to apply the supernode technique. In this method, the two
adjacent nodes that are connected by a voltage source are reduced to a single node and then the equations
are formed by applying Kirchhoff ’s current law as usual. This is explained with the help of Fig. 2.44.
Fig. 2.44
It is clear from Fig. 2.44, that the node 4 is the reference node. Applying Kirchhoff’s current law at the
node 1, we get
I=
V1 V1 −V2
+
R1
R2
Due to the presence of voltage source Vx in between nodes 2 and 3, it is slightly difficult to find out the
current. The supernode technique can be conveniently applied in this case.
Accordingly, we can write the combined equation for nodes 2 and 3 as under.
V2 −V1 V2 V3 −Vy V3
+ +
+ =0
R2
R3
R4
R5
84
Circuits and Networks
The other equation is
V2 – V3 5 Vx
From the above three equations, we can find the three unknown voltages.
EXAMPLE 2.15
Determine the current in the 5 V resistor for the circuit shown in Fig. 2.45.
Fig. 2.45
Solution‰ At the node 1,
10 =
V1 V1 −V2
+
3
2
1 1 V
or V1  +  − 2 −10 = 0
 3 2  2
0.83 V1 – 0.5 V2 – 10 5 0
(2.49)
At nodes 2 and 3, the supernode equation is
V2 −V1 V2 V3 −10 V3
+ +
+ =0
2
1
5
2
or
1 
1 1
−V1
+ V2  + 1 + V3  +  = 2
 2 
 5 2 
2
– 0.5 V1 1 1.5 V2 1 0.7 V3 – 2 5 0
(2.50)
The voltage between nodes 2 and 3 is given by
V2 – V3 5 20
(2.51)
V −10
The current in the 5 V resistor I 5 = 3
5
Solving Eqs (2.49), (2.50), and (2.51), we obtain
V3 5 –8.42 V
−8.42 −10
∴ Current I 5 =
= −3.68 A (current towards node 3) i.e. the current flows towards the node 3.
5
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 6
rr 2-6.1
In the circuit shown in Fig. Q.1, use nodal analysis to find out the voltage across 40 V and the
power supplied by the 5 A source.
Methods of Analysing Circuits
85
Fig. Q.1
rrr 2-6.2
Use nodal analysis in the circuit shown in Fig. Q.2 and determine what value of V will cause V1050.
Fig. Q.2
rrr 2-6.3
Find the value of V0 in the network shown in Fig. Q.3, using PSpice.
Fig. Q.3
rrr 2-6.4
Find the total power dissipation in the following circuit of Fig. Q.4, using PSpice.
Fig. Q.4
86
rrr 2-6.5
Circuits and Networks
For the circuit shown in Fig. Q.5 find the currents in all branches of the circuit using the node
voltage method.
Fig. Q.5
rrr 2-6.6
Write nodal equations for the circuit shown in Fig. Q.6, and find the Voltage V1.
Fig. Q.6
rrr 2-6.7
Use nodal analysis to find V2 in the circuit shown in Fig. Q.7.
Fig. Q.7
Frequently Asked Questions linked to LO 6
rrr 2-6.1
Determine the voltage at each node of the circuit shown in Fig. Q.1.
Fig. Q.1
87
Methods of Analysing Circuits
rrr2-6.2
Using nodal analysis techniques, determine the current 'i' in the network shown in Fig. Q.2.
(JNTU Nov. 2012)
6W
i
3i
3A
10 W
5A
4W
Fig. Q.2
rrr 2-6.3
Using graph theory, find node voltages at (A) and (B) for the network shown in Fig. Q.3.
(PTU 2011-12)
Fig. Q.3
rrr 2-6.4
Two batteries having e/m/f of 10 V and 7 V and internal resistance of 2 W and 3 W respectively are
connected across a load resistance of 1 W. Calculate
(a) Individual battery currents.
(b) Current through the load.
(c) Voltage across the load using nodal analysis.
(PTU 2011-12)
rrr 2-6.5 Using nodal analysis, find 'I' in the circuit shown in Fig. Q.5
(PU 2012)
4W
60
60
60
12V
+
2A
120
I
120
Fig. Q.5
rrr 2-6.6
Find out the current in the 5 W resistance using node voltage analysis and verify the result using
mesh analysis.
(RTU Feb. 2011)
Fig. Q.6
88
rrr2-6.7
Circuits and Networks
For the circuit shown in Fig. Q.7 find node voltage V1, V2, and V3.
(PU 2012)
6V
–+
V1
1W
V2
+ –
0.5 W
6A
1
W
3
0.2 W
V3
3 V1
1W
Fig. Q.7
rrr 2-6.8
rrr 2-6.9
Explain the concept of supernode and supermesh.
(PU 2012)
Find the current in the 6 W resistance in Fig. Q.9 using mesh analysis and verify the result using
nodal analysis.
(RTU Feb. 2011)
Fig. Q.9
2.15
SOURCE TRANSFORMATION TECHNIQUE
In solving networks to find solutions, one may have to deal with energy
LO 7
sources. It has already been discussed in Chapter 1 that basically, energy
sources are either voltage sources or current sources. Sometimes it is
necessary to convert a voltage source to a current source and vice-versa.
Any practical voltage source consists of an ideal voltage source in series
with an internal resistance. Similarly, a practical current source consists of
an ideal current source in parallel with an internal resistance as shown in Fig. 2.46. Rv and Ri represent the
internal resistances of the voltage source Vs, and current source Is, respectively.
Fig. 2.46
Methods of Analysing Circuits
89
Any source, be it a current source or a voltage source, drives current through its load resistance, and
the magnitude of the current depends on the value of the load resistance. Figure 2.47 represents a practical
voltage source and a practical current source connected to the same load resistance RL.
Fig. 2.47
From Fig. 2.47 (a), the load voltage can be calculated by using Kirchhoff’s voltage law as
Vab 5 Vs – IL Rv
The open-circuit voltage VOC 5 Vs
The short-circuit current I SC =
Vs
Rv
From Fig. 2.47 (b),
IL = Is − I = IS −
Vab
RI
The open-circuit voltage VOC 5 IS RI
The short-circuit current ISC 5 IS
The above two sources are said to be equal, if they produce equal amounts of current and voltage when
they are connected to identical load resistances. Therefore, by equating the open-circuit voltages and shortcircuit currents of the above two sources, we obtain
VOC 5 Is RI 5 VS
I SC = I S =
Vs
Rv
It follows that R1 5 RV 5 Rs ∴ Vs 5 IS RS
where RS is the internal resistance of the voltage or current source. Therefore, any practical voltage source,
having an ideal voltage VS and internal series resistance RS can be replaced by a current source IS 5 VS /RS in
parallel with an internal resistance RS. The reverse transformation is also possible. Thus, a practical current
source in parallel with an internal resistance RS can be replaced by a voltage source VS 5 Is Rs in series with
an internal resistance RS.
90
Circuits and Networks
EXAMPLE 2.16
Determine the equivalent voltage source for the current source
shown in Fig. 2.48.
Fig. 2.48
Solution‰ The voltage across terminals A and B is equal to 25
V. Since the internal resistance for the current source is 5 V, the
internal resistance of the voltage source is also 5 V. The equivalent
voltage source is shown in Fig. 2.49.
Fig. 2.49
EXAMPLE 2.17
Determine the equivalent current source for the voltage source shown in
Fig. 2.50.
Fig. 2.50
Solution‰ The short-circuit current at terminals A and B is equal to
I=
50
=1.66 A
30
Since the internal resistance for the voltage source is 30 V, the internal
resistance of the current source is also 30 V. The equivalent current source
is shown in Fig. 2.51.
Fig. 2.51
Frequently Asked Questions linked to LO 7
r 2-7.1
Explain the source-transformation technique.
(AU May/June 2013)
91
Methods of Analysing Circuits
r 2-7.2
Using source transformation replace the current source in the circuit shown in Fig. Q.2 by a voltage
source and find the current delivered by the 50 V voltage source.
(AU Nov./Dec. 2012)
rr 2-7.3 Use source transformation to find I0 in the circuit shown in Fig. Q.3.
(AU April/May 2011)
4W
5W
+
– 50 V
3W
10 A
+
12 V
–
+
10 V –
2W
I0
6A
6W
–
3 W 24 V +
Fig. Q.3
Fig. Q.2
rr 2-7.4
2W
Explain various source-transformation techniques: Using source-transformation techniques, find
current I in the network shown in Fig. Q.4.
(GTU Dec. 2012)
2W
1W
1W
1
2W
2V
+
–
1
2W
1A
1W +
–
2V
2W
Fig. Q.4
r 2-7.5
Explain the source-transformation techniques with
suitable circuits.
(JNTU Nov. 2012)
rr 2-7.6 Obtain Vx using some shifting and source
transformation technique. (Fig. Q.6) (MU 2014)
r 2-7.7 Discuss the properties of an ideal current source
and an ideal voltage source. Explain how a voltage
source can be converted into an equivalent current
source and vice versa.
(RGTU Dec. 2013)
4W
5W
3A
2W
+ V x–
2 Vx
3W
2W
3W
Fig. Q.6
Additional Solved Problems
PROBLEM 2.1
Determine the currents in bridge circuit by using mesh
analysis in Fig. 2.52.
Fig. 2.52
92
Circuits and Networks
Solution‰ Consider ia, ib, and ic are three loop currents as
shown in Fig. 2.53.
By inspection method, we can find the loop equations
3ia2ib2ic 5 10
(2.52)
2ia 1 3ib2ic 5 5
(2.53)
2ia2ib 1 3ic 5 5
(2.54)
From the above equations, we have
 3 −1 −1 ia  10

   
−1
 ib  =  5
3
−
1

   
−1 −1
3  ic   5

∆
The current ia = 1
∆
10 −1 −1


where ∆1 =  5
3 −1 = 120
 5 −1
3

 3 −1 −1


3 −1 = 16
and ∆ = −1
−1 −1
3

∴ ia = ∆1 = 120 = 7.5 A
∆
16
∆2
The current ib =
∆
 3 10 −1 


where ∆2 = −1
5 −1  = 100
−1
5
3 

∴ ib = 100 = 6.25 A
16
∆3
The current ic =
∆
 3 −1 10


where ∆3 = −1
3 5 = 100
−1 −1 5


100
∴ i =
= 6.25 A
c
16
From the loop currents ia, ib, and ic, we can determine the branch currents.
The branch current I 5 ia 5 7.5 A
The branch current I1 5 ia 2 ib 5 1.25 A
The branch current I2 5 ib 5 6.25 A
The branch current I3 5 ib 2 ic 5 0 A
The branch current I4 5 ic 5 6.25 A
The branch current I5 5 ia 2 ic 5 1.25 A
Fig. 2.53
Methods of Analysing Circuits
93
PROBLEM 2.2
For the circuit shown in Fig. 2.54, use nodal analysis to find the current delivered by the 24 V source.
Fig. 2.54
Solution‰ The circuit shown in Fig. 2.54 is a single-node circuit. Consider the node voltage V1 across each
branch in the circuit.
Applying Kirchhoff’s current law at the node, we have
V1 − 24
V
V − 36
−2+ 1 + 1
=0
5
20
10
1 1
1
V1  + +  = 4.8 + 2 + 3.6
 5 20 10 
(2.55)
∴ V1 = 29.7 volts
The current delivered by the 24 V source is
V − 24
I 24 = 1
= 1.14 A
5
PROBLEM 2.3
Determine the current I in the circuit by using loop analysis in Fig. 2.55.
Fig. 2.55
Solution‰ Consider the loop currents as shown in Fig. 2.56.
Fig. 2.56
94
Circuits and Networks
From Fig. 2.56, the supermesh equation can be written as
2(I2 2 I1) 1 2(I2 2 I4) 1 10(I3 2 I1) 1 1.5(I3 2 I5) 5 0
(2.56)
The other equations are
9I4 2 2I2 2 I5 5 0
(2.57)
10.5I5 2 1.5I3 2 I4 5 0
(2.58)
I3 2 I2 5 30
(2.59)
I1 5 20
(2.60)
The above equations can be reduced as
15.5I3 2 2I4 2 1.5I5 5 360
22I3 1 9I4 2 I5 5 2 60
21.5I3 2 I4 1 10.5I5 5 0
From the above equation, we have
 15.5 −2 −1.5  I 3   360 

  

 −2 +9
−1  I 4  = −60

−1.5 −1 10.5  I   0 

  5 

The branch current I 5 I3 2 I5
The loop current I 3 =
∆3
∆
15.5 −2 −1.5
where, ∆ = −2 +9
−1 = 1381
−1.5 −1 10.5
360 −2 −1.5
and
∆3 = −60
9
0 −1
∴ the current I 3 =
−1 = 32310
10.5
∆3 32310
=
= 23.4A
∆
1381
The loop current I 5 =
∆5
∆
15.5 −2 360
where, ∆5 = −2
9 −60 = 4470
−1.5 −1
0
∴ the current I 5 =
∆5 4470
=
= 3.24 A
1381
∆
The branch current I 5 I3 2 I5 5 23.4 2 3.24 5 20.16 A
(2.61)
(2.62)
(2.63)
95
Methods of Analysing Circuits
PROBLEM 2.4
Write nodal equations for the circuit shown in Fig. 2.57 and find the power supplied by the 10 V source.
Fig. 2.57
Solution‰ Consider the nodes a, b and node voltages Va and Vb as shown in Fig. 2.58.
Fig. 2.58
The nodal equation at the node a,
V
V −Vb Va + 10 −Vb
10 + a + a
+
=0
2
2
4
From the above equation, we have
1.25Va 2 0.75Vb 5 2 12.5
The nodal equation at the node b,
Vb −10 −Va Vb −Va
+
− 4V3 + Vb = 0
4
2
From the above equation, we have
2 4.75Va 1 5.75Vb 5 42.5
Since V3 5 Va 1 10 2 Vb
By solving the equations (2.65) and (2.67), we have
Va 5 2 11.03 volts; Vb 5 2 1.724 volts
The current delivered by the 10 V source is I10.
I10 =
Va −Vb + 10
4
(2.64)
(2.65)
(2.66)
(2.67)
96
Circuits and Networks
The power supplied by the 10 V source
V −Vb + 10 
P10 = (10) I10 = 10  a

4


P10 5 1.735 watts.
PROBLEM 2.5
Use mesh analysis to find Vx in the circuit shown in Fig. 2.59.
Fig. 2.59
Solution‰ Consider the loop currents I1, I2, and I3 as shown in Fig. 2.60.
Fig. 2.60
From Fig. 2.60, the loop equations are
30 1 Vx 1 33.33(I1 2 I3) 1 25(I1 2 I2) 1 10 5 0
210 1 25(I2 2 I1) 2 2Vx 5 0
I3 5 0.45 A
(2.68)
(2.69)
and Vx 5 16.67I1
By substituting I3 and Vx and simplifying Eqs (2.68) and (2.69), we get
75I1 2 25I2 5 25
and 2 58.35I1 1 25I2 5 10
By solving Eqs (2.70) and (2.71), we get
I1 5 2.1 A
The voltage across the 16.67 Ω resistor is
Vx 5 I1(16.67) 5 (2.1)(16.67)
Vx 5 35 volts
(2.70)
(2.71)
Methods of Analysing Circuits
97
PROBLEM 2.6
Determine the voltage ratio Vout /Vin for the circuit shown in Fig. 2.61 by using nodal analysis.
Fig. 2.61
Solution‰ I10 1 I3 1 I11 5 0
VA −Vin
10
VA
I3 =
3
V
VA
I11 = , or out
6
I10 =
VA −Vin VA VA
+ + =0
10
3 11
VA Vout
Also 11 = 6
∴ VA 5 Vout 3 1.83
From the above equations, Vout / Vin 5 1/9.53 5 0.105
PROBLEM 2.7
Find the voltages V in the circuit shown in Fig. 2.62 which makes the current in the 10 V resistor zero by
using nodal analysis.
Fig. 2.62
Solution‰ In the circuit shown, assume voltages V1 and V2 at nodes 1 and 2.
At the node 1, the current equation in Fig. 2.63 (a) is
V1 −V V1 V1 −V2
+ +
=0
3
2
10
98
Circuits and Networks
Fig. 2.63 (a)
or
0.93 V1 – 0.1 V2 5 V/3
At the node 2, the current equation in Fig. 2.63 (b) is
V2 −V1 V2 V2 − 50
+ +
=0
10
5
7
or – 0.1 V1 1 0.443 V2 5 7.143
Fig. 2.63 (b)
Since the current in the 10 V resistor is zero, the voltage at the node 1 is equal to the voltage at the node 2.
∴
V1 – V2 5 0
From the above three equations, we can solve for V.
V1 5 20.83 volts and V2 5 20.83 volts
∴ V 5 51.87 V
PROBLEM 2.8
Use nodal analysis to find the power dissipated in the 6 V resistor for the circuit shown in Fig. 2.64.
Fig. 2.64
Methods of Analysing Circuits
99
Solution‰ Assume voltage V1, V2 , and V3 at nodes 1, 2, and 3 as shown in Fig. 2.64.
By applying current law at the node 1, we have
V1 − 20 V1 −V2 V1 −V3
+
+
=0
3
1
2
or 1.83V1 – V2 – 0.5V3 5 6.67
(2.72)
At the node 2,
V2 −V1 V2 −V3
+
=5 A
1
6
or – V1 – 1.167V2 – 0.167V3 5 5
(2.73)
At the node 3,
V3 −V1 V3 −V2 V3
+ =0
+
2
6
5
or – 0.5 V1 – 0.167 V2 1 0.867 V3 5 0
Applying Cramer’s rule to Eqs (2.72), (2.73) and (2.74), we have
V2 =
∆2
∆
1.83 −1
−0.5 


where ∆ = −1
−1.167 −0.167 = −2.64

−0.5 −0.167 0.867 


1.83 6.67 −0.5 


∆2 = −1
5
−0.167 = 13.02
−0.5 0
0.867 

∴
V2 =
13.02
= −4.93 V
−2.64
V2 5 – 4.93 V
Similarly,
V3 =
∆3
∆
1.83 −1
6.67


∆3 = −1
−1.167 5  = 1.25
−0.5 −0.167 0 


∴
V3 =
1.25
=− 0.47 V
−2.64
(2.74)
100
Circuits and Networks
The current in the 6 V resistor is
I6 =
V2 −V3
6
−4.93 + 0.47
=− 0.74 A
6
The power absorbed or dissipated 5 I 62 R6
=
5 (0.74)2 3 6
5 3.29 W
PROBLEM 2.9
Determine the power dissipated by 5 V resistor in the circuit shown in Fig. 2.65.
Fig. 2.65
Solution‰ In Fig. 2.65, assume voltages V1, V2, and V3 at nodes 1, 2, and 3. At the node 1, the current law gives
V1 − 40 −V3 V1 −V2
+
− 3− 5 = 0
4
6
or 0.42 V1 – 0.167 V2 – 0.25 V3 5 18
Applying the supernode technique between nodes 2 and 3, the combined equation at node 2 and 3 is
V2 −V1
V V V + 40 −V1
+5+ 2 + 3 + 3
=0
6
3
5
4
or – 0.42 V1 1 0.5 V2 1 0.45 V3 5 –15
Also, V3 – V2 5 20 V
Solving the above three equations, we get
V1 5 52.89 V, V2 5 – 1.89 V and
V3 5 18.11 V
∴ the current in the 5V resistor I 5 =
V3
5
Methods of Analysing Circuits
101
18.11
= 3.62 A
5
The power absorbed by the 5 V resistor P5 5 I 52 R5
5 (3.62)2 3 5
5 65.52 W
PROBLEM 2.10
Find the power delivered by the 5 A current source in the circuit shown in Fig. 2.66 by using the nodal method.
Fig. 2.66
Solution‰ Assume the voltages V1, V2, and V3 at nodes 1, 2, and 3, respectively. Here, the 10 V source is
common between nodes 1 and 2. So applying the supernode technique, the combined equation at nodes 1
and 2 is
V1 −V3
V −V3
V
+2+ 2
−5 + 2 = 0
3
1
5
or 0.34 V1 1 1.2 V2 – 1.34 V3 5 3
At the node 3,
V3 −V1 V3 −V2 V3
+ =0
+
3
1
2
or – 0.34 V1 – V2 1 1.83 V3 5 0
Also, V1 – V2 5 10
Solving the above equations, we get
V1 5 13.72 V; V2 5 3.72 V
V3 5 4.567 V
Hence, the power delivered by the source (5 A) 5 V2 3 5
5 3.72 3 5 5 18.6 W
PROBLEM 2.11
Using source transformation, find the power delivered by the 50 V voltage source in the circuit shown in Fig.
2.67.
102
Circuits and Networks
Fig. 2.67
Solution‰ The current source in the circuit in Fig. 2.67 can be replaced by a voltage source as shown in Fig.
2.68.
V − 50 V − 20 V −10
+
+
=0
5
2
3
V [0.2 1 0.5 1 0.33] 5 23.33
or V =
23.33
= 22.65 V
1.03
∴ the current delivered by the 50 V voltage source
is (50 – V )/5
50 − 22.65
= 5.47 A
5
Hence, the power delivered by the 50 V voltage source 5 50 3 5.47 5 273.5 W
Fig. 2.68
PROBLEM 2.12
By using source transformation, source combination and resistance combination convert the circuit shown in
Fig. 2.69 into a single voltage source and single resistance.
Fig. 2.69
Solution‰ The voltage source in the circuit of Fig. 2.69 can be replaced by a current source as shown in Fig.
2.70 (a).
103
Methods of Analysing Circuits
Fig. 2.70 (a)
Here, the current sources can be combined into a single source. Similarly, all the resistances can be
combined into a single resistance, as shown in Fig. 2.70 (b).
Figure 2.70 (b) can be replaced by single voltage source and a series resistance as shown in Fig. 2.70 (c).
Fig. 2.70 (b)
Fig. 2.70 (c)
PROBLEM 2.13
For the circuit shown in Fig. 2.71 find the voltage across the 4 V resistor by using nodal analysis.
Fig. 2.71
Solution‰ In the circuit shown, assume voltages V1 and V2 at nodes 1 and 2. At the node 1, the current equation is
5+
V1 V1 + 5 −V2 V1 −V2
+
+
=0
3
4
2
or
1.08 V1 – 0.75 V2 5 – 6.25
At the node 2, the current equation is
V2 −V1 − 5 V2 −V1
V
+
− 4Vx + 2 = 0
4
2
1
Vx 5 V1 1 5 – V2
(2.75)
104
or
Circuits and Networks
– 4.75 V1 1 5.75 V2 5 21.25
(2.76)
Applying Cramer’s rule to Eqs (2.75) and (2.76), we have
∆
V2 = 2
∆
1.08 −0.75
where
∆=
= 2.65
−4.75
5.75
∆2 =
1.08 −6.25
= −6.74
−4.75 21.25
∴ V2 =
∆2 −6.74
=
= −2.54 V
∆
2.65
Similarly,
∆1
∆
−6.25 −0.75
∆1 =
= −20
21.25
5.75
V1 =
V1 =
∆1 −20
=
= −7.55 V
∆
2.65
The voltage across the 4 V resistor is
Vx 5 V1 1 5 – V2
5 – 7.55 1 5 – (– 2.54)
Vx 5 – 0.01 volts
PROBLEM 2.14
For the circuit shown in Fig. 2.72, find the current passing through the 5 V resistor by using the nodal method.
Fig. 2.72
Solution‰ In the circuit shown, assume the voltage V at the node 1.
At the node 1, the current equation is
where
V − 36 − 6 I1
V − 30
−2+
=0
6
5
V − 30
I1 =
5
105
Methods of Analysing Circuits
From the above equation
V 5 48 V
The current in the 5 V resistor is
V − 30
I1 =
= 3.6 A
5
PROBLEM 2.15
In the circuit shown in Fig. 2.73, find the power
delivered by the 4 V source using mesh analysis and
voltage across the 2 V resistor.
Solution‰ Since branches BC and DE consist of
current sources, we use the supermesh technique.
The combined supermesh equation is
2I1 1 6I1 1 4(I1 – I3) 1 (I2 – I3) – 4 1 5I2 5 0
or
12I1 1 6I2– 5I3 5 4
In the branch BC, I2 – I1 5 5
V
In the branch DE, I 3 = 2
2
Solving the above equations,
Fig. 2.73
I1 5 – 2 A; I2 5 3 A
The voltage across the 2 V resistor V2 5 2I1 5 2 3 (–2) 5 – 4 V
Power delivered by the 4 V source P4 5 4I2 5 4(3) 5 12 W
PROBLEM 2.16
For the circuit shown in Fig. 2.74, find the current through the 10 V resistor by using mesh analysis.
Fig. 2.74
Solution‰ The parallel branches consist of current sources. Here, we use supermesh analysis. The combined
supermesh equation is.
or – 15 1 10I1 1 20 1 5I2 1 4I3 – 40 5 0
and 10I1 1 5I2 1 4I3 5 35
I1 – I2 5 2
106
Circuits and Networks
I3 – I2 5 2I1
Solving the above equations, we get
I1 5 1.96 A
The current in the 10 V resistor is I1 5 1.96 A
PSpice Problems
PROBLEM 2.1
Determine the currents for the circuit shown in Fig. 2.75.
* CURRENTS WITH DC ANALYSIS
V1 2 1 DC 10 V
Fig. 2.75
V2 5 4 DC 20 V
V3 7 6 DC 30 V
I1 0 3 5 A
I2 5 0 10 A
I3 7 8 15 A
R1 2 3
1
R2 3 4 4
R3 5 6 8
R4 7 0 20
R5 8 1 16
R6 1 0 10
.DC LIN V1 10 10 1
.PRINT DC I(R1) I(R2) I(R3) I(R5)
.END
**** DC TRANSFER CURVES TEMPERATURE 5 27.000 DEG C
****************************************************************
V1
I(R1)
I(R2)
I(R3)
I(R5)
1.000 E 1 01 1.465 E 1 01 1.965 E 1 01 9.651 E 1 00 1.500 E 1 01
Result
I1 5 I(R1) 5 14.65 A
I2 5 I(R2) 5 19.65 A
I3 5 I(R5) 5 15 A
I4 5 I(R3) 5 9.65 A
Methods of Analysing Circuits
PROBLEM 2.2
Determine the power delivered by the 5 A current source in
the circuit shown in Fig. 2.76.
* POWER DELIVERED BY DC SOURCE
I1 1 0 2 A
V1 1 2 10 V
I2 0 2 5 A
R1 2 3 1
R2 1 3 3
Fig. 2.76
R3 3 0 2
R4 2 0 5
.DC LIN V1 10 10 1
.PRINT DC V(1) V(2) V(3)
.END
**** DC TRANSFER CURVES TEMPERATURE 5 27.000 DEG C
****************************************************************
V1
V(1)
V(2)
V(3)
1.000 E 1 01
1.371 E 1 01
3.710 E 1 00
4.516 E 1 00
Result
Power delivered by the 5 A source = V2 X 5 5 3.71 X 5 5 18.55 W
PROBLEM 2.3
Determine I in the Fig. 2.77 using PSpice, with I1 varying for 5 A, 10 A and 20 A.
Fig. 2.77
* TO DETERMINE CURRENT IN AN ELEMENT USING DC SWEEP
I1 0 1 20 A
I2 2 3 30
R1 1 2 2
R2 1 3 2
R3 1 4 6
R4 2 0 10
R5 3 0 1.5
R6 4 0 8
R7 3 4 1
107
108
Circuits and Networks
.DC LIN I1 LIST 5 10 20
.PRINT DC I(R5)
.END
OUTPUT
**** DC TRANSFER CURVES TEMPERATURE 5 27.000 DEG C
****************************************************************
I1
I(R5)
5.000 E 1 00
9.993 E 1 00
1.000 E 1 01
1.338 E 1 01
2.000 E 1 01
2.016 E 1 01
Result
I 5 I (R5) 5 20.16 A
Answers to Practice Problems
2-5.1
2-5.3
2-5.4
2-5.6
2-5.7
2-5.8
2-6.1
2-6.5
2580 W; –32 V
R1 5 4.88 Ω;
R2 5 3.82 Ω;
2.65 V
1.2 A; 4.2 A; 2 A; 3.2 A
I 5 11A
I1 5 5 A; I2 5 –11 A
2-6.6
–60.9 V; 195.7 W
I1 5 – 0.92A, I2 5 0.615 A
I3 5 1.075A, I4 5 1.69 A
V1 5 148.1 volts
Objective-Type Questions
rrr 2.1
A tree has
rrr 2.2
The number of branches in a tree is _____ the number of branches in a graph.
(a) less than
(b) more than
(c) equal to
rrr 2.3
The tie-set schedule gives the relation between
(a) branch currents and link currents
(c) branch currents and link voltages
(b) branch voltages and link currents
(d) none of the above
The cut-set schedule gives the relation between
(a) branch currents and link currents
(c) branch voltages and link voltages
(b) branch voltages and tree branch voltages
(d) branch current and tree currents
Mesh analysis is based on
(a) Kirchhoff’s current law
(c) both
(b) Kirchhoff’s voltage law
(d) none
(a) a closed path
rrr 2.4
rrr 2.5
rrr 2.6
rrr 2.7
(b) no closed paths
(c) none
If a network contains B branches, and N nodes, then the number of mesh current equations would be
(a) B – (N – 1)
(b) N – (B – 1)
(c) B – N – 1
(d) (B 1 N ) – 1
A network has 10 nodes and 17 branches. The number of different node pair voltages would be
(a) 7
(b) 9
(c) 45
(d) 10
rrr 2.8 A practical voltage source consists of
109
Methods of Analysing Circuits
(a) an ideal voltage source in series with an internal resistance
(b) an ideal voltage source in parallel with an internal resistance
(c) both (a) and (b) are correct
(d) none of the above
rrr 2.9 A practical current source consists of
(a) an ideal current source in series with a resistance
(b) an ideal current source in parallel with a resistance
(c) both are correct
(d) none of the above
rrr 2.10 A circuit consists of two resistances, R1 and R2, in parallel. The total current passing through the circuit is IT.
The current passing through R1 is
(a)
IT R1
R1 + R2
(b)
IT ( R1 + R2 )
R1
(c)
IT R2
R1 + R2
(d)
IT R1 + R2
R2
rrr 2.11 A network has seven nodes and five independent loops. The number of branches in the network is
(a) 13
(b) 12
(c) 11
(d) 10
rrr 2.12 The nodal method of circuit analysis is based on
(a) KVL and Ohm’s law
(b) KCL and Ohm’s law
(c) KCL and KVL
(d) KCL, KVL, and Ohm’s law
rrr 2.13 The number of independent loops for a network with n nodes and b branches is
(a) n – 1
(b) b – n
(c) b – n 1 1
(d) independent of the number of nodes
rrr 2.14 The two electrical subnetworks N1 and N2 are connected through three resistors as shown in Fig. 2.78. The
voltage across the 5 V resistor and the 1 V resistor are given to be 10 V and 5 V respectively. The voltage
across the 15 V resistor is
(a) – 105 V
(b) 1 105 V
(c) – 15 V
(d) 1 15 V
Fig. 2.78
rrr 2.15 Relative to a given fixed tree of a network
(a) link currents form an independent set
(c) link voltages form an independent set
(b) branch currents form an independent set
(d) branch voltages form an independent set
For interactive quiz with answers,
scan the QR code given here
OR
visit
http://qrcode.flipick.com/index.php/260
3
LEARNING OBJECTIVES
3.1
STAR-DELTA TRANSFORMATION
In the preceding chapter, a simple technique called the source transformation
technique was discussed. The star-delta transformation is another technique useful
in solving complex networks. Basically, any three circuit elements, i.e. resistive,
inductive or capacitive, may be
connected in two different ways. One
way of connecting these elements
is called the star connection, or the
Y-connection. The other way of
connecting these elements is called
the delta (D) connection. The circuit is
said to be in star connection, if three
elements are connected as shown in
Fig. 3.1 (a), when it appears like a star
(Y ). Similarly, the circuit is said to be
in delta connection, if three elements
are connected as shown in Fig. 3.1
(b), when it appears like a delta (D).
Fig. 3.1
LO 1
Useful Theorems in Circuit Analysis
111
The above two circuits are equal if their respective resistances from the terminals AB, BC, and CA are
equal. Consider the star-connected circuit in Fig. 3.1 (a); the resistance from the terminals AB, BC, and CA
respectively are
RAB (Y ) 5 RA 1 RB
RBC (Y ) 5 RB 1 RC
RCA (Y ) 5 RC1 RA
Similarly, in the delta-connected network in Fig. 3.1 (b), the resistances seen from the terminals AB, BC,
and CA respectively are
RAB (∆) = R1 ( R2 + R3 ) =
R1 ( R2 + R3 )
R1 + R2 + R3
RBC (∆) = R3 ( R1 + R2 ) =
R3 ( R1 + R2 )
R1 + R2 + R3
RCA (∆) = R2 ( R1 + R3 ) =
R2 ( R1 + R3 )
R1 + R2 + R3
Now, if we equate the resistances of star and delta circuits, we get
RA + RB =
R1 ( R2 + R3 )
R1 + R2 + R3
(3.1)
RB + RC =
R3 ( R1 + R2 )
R1 + R2 + R3
(3.2)
RC + RA =
R2 ( R1 + R3 )
R1 + R2 + R3
(3.3)
Subtracting Eq. (3.2) from Eq. (3.1), and adding Eq. (3.3) to the resultant, we have
RA =
R1 R2
R1 + R2 + R3
(3.4)
Similarly,
RB =
R1 R3
R1 + R2 + R3
(3.5)
and
RC =
R2 R3
R1 + R2 + R3
(3.6)
Thus, a delta connection of R1, R2, and R3 may be replaced by a star connection of RA, RB, and RC as
determined from Eqs (3.4), (3.5) and (3.6). Now if we multiply the Eqs (3.4) and (3.5), (3.5) and (3.6), (3.6)
and (3.4), and add the three, we get the final equation as under:
R 2 R R + R32 R1 R2 + R22 R1 R3
RA RB + RB RC + RC RA = 1 2 3
(3.7)
( R1 + R2 + R3 ) 2
In Eq. (3.7) dividing the LHS by RA, gives R3; dividing it by RB gives R2, and doing the same with RC ,
gives R1.
112
Circuits and Networks
Thus R1 =
RA RB + RB RC + RC RA
RC
R2 =
RA RB + RB RC + RC RA
RB
RA RB + RB RC + RC RA
RA
From the above results, we can say that a star-connected
circuit can be transformed into a delta-connected circuit and
vice-versa.
From Fig. 3.2 and the above results, we can conclude that any
resistance of the delta circuit is equal to the sum of the products
of all possible pairs of star resistances divided by the opposite
resistance of the star circuit. Similarly, any resistance of the star
circuit is equal to the product of two adjacent resistances in the
delta-connected circuit divided by the sum of all resistances in
delta-connected circuit.
R3 =
and
Fig. 3.2
EXAMPLE 3.1
Obtain the star-connected equivalent for the deltaconnected circuit shown in Fig. 3.3.
Solution The above circuit can be replaced by a star-
connected circuit as shown in Fig. 3.4 (a).
Fig. 3.3
Performing the D to Y transformation, we obtain
13×12
13×14
, R2 =
14 + 13 + 12
14 + 13 + 12
14 ×12
R3 =
14 + 13 + 12
R1 =
and
Fig. 3.4
Useful Theorems in Circuit Analysis
∴
R1 5 4 V, R2 5 4.66 V, R3 5 4.31 V
The star-connected equivalent is shown in Fig. 3.4 (b).
EXAMPLE 3.2
Obtain the delta-connected equivalent for the star-connected circuit shown in Fig. 3.5.
Fig. 3.5
Solution The above circuit can be replaced by a delta-connected circuit as shown in Fig. 3.6 (a).
Performing the Y to D transformation, we get from Fig. 3.6 (a),
and
R1 =
20 ×10 + 20 ×5 + 10 ×5
= 17.5 V
20
R2 =
20 ×10 + 20 ×5 + 10 ×5
= 35 V
10
R3 =
20 ×10 + 20 ×5 + 10 ×5
= 70 V
5
Fig. 3.6
The equivalent delta circuit is shown in Fig. 3.6 (b).
113
114
Circuits and Networks
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 1*
For the bridge network shown in Fig. Q.2 (a), determine the total resistance seen from terminals
AB by using star-delta transformation.
rrr3-1.2 Find the equivalent resistance across the terminals A and B of the network shown in Fig. Q.2 (b)
using star-delta transformation.
rrr 3-1.1
20 V
Fig. Q.2
rrr 3-1.3
2.2 k
Determine the voltages and currents
of the resistances in the circuit
shown in Fig. Q.3 using source
transformation technique.
6.8 k
Fig. Q.3
Frequently Asked Questions linked to LO 1*
rrr 3-1.1
Derive the expression for star-connected resistances in terms of delta-connected resistances.
[AU May/June 2013]
rrr 3-1.2 Give a delta circuit it having resistors, write the required expressions to transform the circuit to a
star circuit.
[AU Nov./Dec. 2012]
rrr 3-1.3 Calculate the equivalent resistance Rab when all the resistance values are equal to 1 W for the circuit
shown below. (Fig. Q.3)
[AU Nov./Dec. 2012]
rrr 3-1.4 Transform the circuit shown below, from D to Y. (Fig. Q. 4)
[AU April/May 2011]
Reg
A
c
a
30 W
e
30 W
30 W
b
Fig. Q.3
d
B
C
Fig. Q.4
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
115
Useful Theorems in Circuit Analysis
rrr 3-1.5
Use the technique of D-Y conversion to find the equivalent resistance between terminals A-B of the
circuit shown below.
[AU April/May 2011]
A
1W
4W
3W
2W
B
5W
Fig. Q.5
rrr 3-1.6
rrr 3-1.7
3.2
State and explain star-delta conversion in ac systems.
Obtain the expressions for star-delta equivalence of an impedance network.
[JNTU Nov. 2012]
[JNTU Nov. 2012]
SUPERPOSITION THEOREM
The superposition theorem states that in any linear network containing
LO 2
two or more sources, the response in any element is equal to the algebraic
sum of the responses caused by individual sources acting alone, while the
other sources are non-operative; that is, while considering the effect of
individual sources, other ideal voltage sources and ideal current sources in
the network are replaced by short circuit and open circuit across their terminals. This theorem is valid
only for linear systems. This theorem can be better understood with a numerical example.
Consider the circuit which contains two sources as shown in Fig. 3.7.
Now let us find the current passing through the 3 V resistor in the circuit. According to the superposition
theorem, the current I2 due to the 20 V voltage source with 5 A source open circuited 5 20/(5 1 3) 5 2.5 A
(see Fig. 3.8)
Fig. 3.7
Fig. 3.8
The current I5 due to the 5 A source with the 20 V source short circuited is
I 5 = 5×
5
= 3.125 A
(3 + 5)
The total current passing through the 3 V resistor is
(2.5 1 3.125) 5 5.625 A
Let us verify the above result by applying nodal analysis.
116
Circuits and Networks
Fig. 3.9
Fig. 3.10
The current passing in the 3 V resistor due to both sources should be 5.625 A.
Applying nodal analysis to Fig. 3.10, we have
V − 20 V
+ =5
5
3
1 1
V  +  = 5+4
 5 3 
15
V = 9× = 16.875 V
8
The current passing through the 3 V resistor is equal to V/3,
16.875
i.e. I =
= 5.625 A
3
So the superposition theorem is verified.
Let us now examine the power responses.
Power dissipated in the 3 V resistor due to the voltage source acting alone
P20 5 (I2)2R 5 (2.5)2 3 5 18.75 W
Power dissipated in the 3 V resistor due to the current source acting alone
P5 5 (I5)2R 5 (3.125)2 3 5 29.29 W
Power dissipated in the 3 V resistor when both the sources are acting simultaneously is given by
P 5 (5.625)2 3 3 5 94.92 W
From the above results, the superposition of P20 and P5 gives
P20 1 P5 5 48.04 W
which is not equal to P 5 94.92 W
We can, therefore, state that the superposition theorem is not valid for power responses. It is applicable
only for computing voltage and current responses.
EXAMPLE 3.3
Find the voltage across the 2 V resistor in Fig. 3.11 by
using the superposition theorem.
Fig. 3.11
Useful Theorems in Circuit Analysis
117
Solution Let us find the voltage across the 2 V resistor due to individual sources. The algebraic sum of these voltages
gives the total voltage across the 2 V resistor.
Our first step is to find the voltage across the 2 V resistor due to the 10 V source, while other sources are
set equal to zero.
The circuit is redrawn as shown in Fig. 3.12 (a).
Fig. 3.12
Assuming a voltage V at the node ‘A’ as shown in Fig. 3.12 (a), the current equation is
V −10 V V
+ + =0
10
20 7
V [0.1 1 0.05 1 0.143] 5 1
or
V 5 3.41 V
The voltage across the 2 V resistor due to the 10 V source is
V2 =
V
× 2 = 0.97 V
7
Our second step is to find out the voltage across the 2 V resistor due to the 20 V source, while the other
sources are set equal to zero. The circuit is redrawn as shown in Fig. 3.12 (b).
Assuming voltage V at the node A as shown in Fig. 3.12 (b), the current equation is
V − 20 V
V
+ + =0
7
20 10
V [0.143 1 0.05 1 0.1] 5 2.86
or V =
2.86
= 9.76 V
0.293
The voltage across the 2 V resistor due to the 20 V source is
V − 20 
× 2 = − 2.92 V
V2 = 
 7 
The last step is to find the voltage across the 2 V resistor
due to the 2 A current source, while the other sources
are set equal to zero. The circuit is redrawn as shown in
Fig. 3.12 (c).
Fig. 3.12 (c)
118
Circuits and Networks
The current in the 2 V resistor = 2×
5
5 + 8.67
10
= 0.73 A
13.67
The voltage across the 2 V resistor 5 0.73 3 2 5 1.46 V
The algebraic sum of these voltages gives the total voltage across the 2 V resistor in the network
V 5 0.97 – 2.92 – 1.46 5 – 3.41 V
The negative sign of the voltage indicates that the voltage at ‘A’ is negative.
=
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 2
rrr 3-2.1
Find the current I in the circuit shown in Fig. Q.1 by
using the superposition theorem.
rrr 3-2.2
Find the current in the 4 V resistor of the circuit
shown in Fig. Q.2 by using the superposition
theorem.
Fig. Q.1
Fig. Q.2
Calculate the new current in the circuit shown in
Fig. Q.3 when the resistor R3 is increased by 30%.
rrr3-2.4 The circuit shown in Fig. Q.4 consists of
dependent source. Use the superposition
theorem to find the current I in the 3 V resistor.
rrr 3-2.5 Obtain the current passing through the 2 V
resistor in the circuit shown in Fig. Q.5 by
using the superposition theorem.
rrr 3-2.3
Fig. Q.4
Fig. Q.3
Fig. Q.5
119
Useful Theorems in Circuit Analysis
rrr 3-2.6
Using PSPICE, determine V0 in the
following circuit, shown in Fig. Q.6, using
the superposition principle. Calculate the
power delivered to the 3 V resistor.
Fig. Q.6
Frequently Asked Questions linked to LO 2
rrr 3-2.1
Calculate the current in the 4
resistor of Fig. Q.1 using the superposition theorem.
[AU May/June 2014]
3W
10 V
2W
2W
4W
1A
Fig. Q.1
rrr 3-2.2
Find the current through various branches of the circuit shown in Fig. Q.2, by employing the
superposition theorem.
[AU Nov./Dec. 2012]
5W
10 W
50 V
10 W
5A
Fig. Q.2
rrr 3-2.3
Determine the voltage across the 20
superposition theorem.
resistance in the circuit shown in Fig. Q.3, using the
[AU April/May 2011]
Fig. Q.3
rrr 3-2.4
Find the voltage across the 1 k
theorem.
resistor in the circuit shown in Fig. Q.4 using the superposition
[GTU Dec. 2010]
1 kW
10 mA
3 kW
15 V
Fig. Q.4
4W
120
Circuits and Networks
rrr 3-2.5
Determine the node voltage V1 and V2 in the network shown in Fig. Q.5 by applying the superposition
theorem.
[GTU Dec. 2010]
1
2 V dc
2
V1
2
V2
1 A dc
2
4 V dc
GND
Fig. Q.5
rrr 3-2.6
State and explain the superposition theorem. Hence, using this, find Vab in Fig. Q.6.
[GTU Dec. 2012]
3A
6W
4W
A
30V
8A
2W
B
Fig. Q.6
rrr 3-2.7
State the superposition theorem. Explain an example.
[PTU 2011-2012]
rrr3-2.8 Find the current through RL = 7.5
using superposition theorem in the network shown in Fig.
Q.8.
[PU 2010]
4W
2W
RL
10 A
5W
2W
20 V
Fig. Q.8
3.3
THEVENIN’S THEOREM
In many practical applications, it is always not necessary to analyse the complete
LO 3
circuit; it requires that the voltage, current, or power in only one resistance of a
circuit be found. The use of this theorem provides a simple, equivalent circuit
which can be substituted for the original network. Thevenin’s theorem states
that any two terminal linear network having a number of voltage current sources
and resistances can be replaced by a simple equivalent circuit consisting of a single voltage source in series with
a resistance, where the value of the voltage source is equal to the open-circuit voltage across the two terminals
of the network, and resistance is equal to the equivalent resistance measured between the terminals with all the
energy sources are replaced by their internal resistances. According to Thevenin’s theorem, an equivalent circuit
can be found to replace the circuit in Fig. 3.13.
121
Useful Theorems in Circuit Analysis
In the circuit, if the 24 V load resistance is connected to
Thevenin’s equivalent circuit, it will have the same current through
it and the same voltage across its terminals as it experienced in
the original circuit. To verify this, let us find the current passing
through the 24 V resistance due to the original circuit.
12
I 24 = IT ×
12 + 24
10
10
Fig. 3.13
where IT =
= =1 A
2 + (12 24) 10
12
∴ I 24 = 1×
= 0.33 A
12 + 24
The voltage across the 24 V resistor 5 0.33 3 24 5 7.92 V.
Now let us find Thevenin’s equivalent circuit.
The Thevenin voltage is equal to the open-circuit voltage across the terminals ‘AB’, i.e. the voltage across
the 12 V resistor. When the load resistance is disconnected from the circuit, the Thevenin voltage
12
VTh =10× = 8.57 V
14
The resistance into the open-circuit terminals is equal to the
Thevenin resistance
RTh =
12×2
=1.71 V
14
Thevenin’s equivalent circuit is shown in Fig. 3.14.
Now let us find the current passing through the 24 V resistance
and voltage across it due to Thevenin’s equivalent circuit.
I 24 =
Fig. 3.14
8.57
= 0.33 A
24 +1.71
The voltage across the 24 V resistance is equal to 7.92 V. Thus, it is proved that RL (5 24 V) has the same
values of current and voltage in both the original circuit and Thevenin’s equivalent circuit.
EXAMPLE 3.4
Determine the Thevenin’s equivalent circuit across ‘AB’ for
the given circuit shown in Fig. 3.15.
Fig. 3.15
Solution The complete circuit can be replaced by a voltage source in series with a resistance as shown in
Fig. 3.16 (a)
where VTh is the voltage across terminals AB, and
RTh is the resistance seen into the terminals AB.
122
Circuits and Networks
To solve for VTh, we have to find the voltage drops around the closed path as shown in Fig. 3.16 (b).
We have
50 – 25 5 10I 1 5I
or 15I 5 25
25
∴ I = =1.67 A
15
Fig. 3.16
Voltage across 10 V 5 16.7 V
Voltage drop across 5 V 5 8.35 V
or VTh 5 VAB 5 50 – V10
5 50 – 16.7 5 33.3 V
To find RTh, the two voltage sources are removed and replaced
with short circuit. The resistance at terminals AB then is the parallel
combination of the 10 V resistor and 5 V resistor; or
10×5
RTh =
= 3.33 V
15
Thevenin’s equivalent circuit is shown in Fig. 3.16 (c).
Fig. 3.16 (c)
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 3
Determine the current passing through the 2 V resistor
by using Thevenin’s theorem in the circuit shown in
Fig. Q.1.
rrr3-3.2 Find Thevenin’s equivalent circuit for the network
shown in Fig. Q.2 and hence, find the current passing
through the 10 V resistor.
rrr 3-3.3 Find the Thevenin equivalent circuit of the circuit
seen from AB using PSPICE (Fig. Q.3).
rrr 3-3.1
Fig. Q.2
Fig. Q.1
Fig. Q.3
123
Useful Theorems in Circuit Analysis
Frequently Asked Questions linked to LO 3
rrr 3-3.1
Why do you short circuit the voltage source and open the current source when you find Thevenin's
voltage of a network?
[AU May/Jnue 2014]
rrr 3-3.2
For the circuit shown in Fig. Q.2 using Thevenin's theorem, the current in the 10-ohm resistor.
[AU May/Jnue 2014]
10 W
10 V
+
–
3W
+
–
3W
4V
1W
2W
Fig. Q.2
rrr 3-3.3
State Thevenin's theorem.
rrr 3-3.4
State and prove Thevenin's theorem, find Rth and Vth for the network shown in Fig. Q.4.
[GTU Dec. 2012]
[AU April/May 2011]
Fig. Q.4
Obtain Thevenin's equivalent circuit (Q.5). Hence, find the current flowing through the 10W load.
[MU 2014]
4W
8W
3 Ix
+
5W
2W
Ix
SW
Vx
7W
8V
4Vx
4V
1W
–
rrr 3-3.5
3W
10 W
6V
6W
Fig. Q.5
rrr 3-3.6
Obtain the Thevenin's equivalent circuit for the network shown in Fig. Q.6.
2 kW
4V
[PTU 2009-10]
3 kW
V1
4000
Vx
Fig. Q.6
rrr 3-3.7
State Thevenin's theorem and find the current through the 4 W resistor using Thevenin's theorem for
the circuit shown in Fig. Q.7.
[PU 2010]
124
Circuits and Networks
15V
15 W
30 W
4W
5W
60 W
Fig. Q.7
rrr 3-3.8
State and prove Thevenin's theorem. Show with an example how theorem can be usefully employed
in circuit analysis.
rrr 3-3.9 Draw the Thevenin's equivalent of the circuit shown in Fig. Q.9.
[RGTU June 2014]
1W
10 V
1W
1W
5V
2W
1W
2W
Fig. Q.9
3.4
NORTON’S THEOREM
Another method of analysing the circuit is given by Norton’s theorem, which
LO 4
states that any two terminal linear network with current sources, voltage
sources and resistances can
be replaced by an equivalent
circuit consisting of a current
source in parallel with a resistance. The value of the current
source is the short-circuit current between the two terminals
of the network and the resistance is the equivalent resistance
measured between the terminals of the network with all the
Fig. 3.17
energy sources are replaced by their internal resistance.
According to Norton’s theorem, an equivalent circuit can
be found to replace the circuit in Fig. 3.17.
In the circuit, if the load resistance of 6 V is connected to Norton’s equivalent circuit, it will have the same
current through it and the same voltage across its terminals as it experiences in the original circuit. To verify
this, let us find the current passing through the 6 V resistor due to the original circuit.
10
10 + 6
20
where IT =
= 2.285 A
5 + (10 6)
10
∴ I 6 = 2.285× = 1.43 A
16
i.e. the voltage across the 6 V resistor is 8.58 V. Now let us find Norton’s equivalent circuit. The magnitude
of the current in the Norton’s equivalent circuit is equal to the current passing through short-circuited terminals
as shown in Fig. 3.18.
I 6 = IT ×
20
=4A
5
Norton’s resistance is equal to the parallel combination of both the 5 V and 10 V resistors.
Here, I N =
125
Useful Theorems in Circuit Analysis
RN =
5×10
= 3.33 V
15
Fig. 3.18
Fig. 3.19
The Norton’s equivalent source is shown in Fig. 3.19.
Now let us find the current passing through the 6 V resistor and the voltage across it due to Norton’s
equivalent circuit.
3.33
I 6 = 4×
=1.43 A
6 + 3.33
The voltage across the 6 V resistor 5 1.43 3 6 5 8.58 V
Thus, it is proved that RL (5 6 V) has the same values of current and voltage in both the original circuit
and Norton’s equivalent circuit.
EXAMPLE 3.5
Determine Norton’s equivalent circuit at terminals AB for the
circuit shown in Fig. 3.20.
Solution The complete circuit can be replaced by a current
source in parallel with a single resistor as shown in Fig. 3.21
Fig. 3.20
(a), where IN is the current passing through the short circuited
output terminals AB and RN is the resistance as seen into the output terminals.
To solve for IN , we have to find the current passing through the terminals AB as shown in Fig. 3.21 (b).
From Fig. 3.21 (b), the current passing through the terminals AB is 4 A. The resistance at terminals AB is
the parallel combination of the 10 V resistor and the 5 V resistor,
10 ×5
or RN =
= 3.33 V
10 + 5
Norton’s equivalent circuit is shown in Fig. 3.21 (c).
Fig. 3.21
126
Circuits and Networks
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 4
rrr 3-4.1
Find the Thevenin’s and Norton’s equivalent circuits across terminals ab for the given circuit
shown in Fig. Q.1.
2 kW
3 kW
a
b
Fig. Q.1
rrr 3-4.2
Obtain Norton’s equivalent circuit of the network shown in
Fig. Q.2.
rrr 3-4.3
Using PSPICE, for the circuit shown in Fig. Q.3 obtain
Norton’s equivalent or viewed from terminals
(a) a – b
(b) c – d
Fig. Q.2
Fig. Q.3
Frequently Asked Questions linked to LO 4
rrr 3-4.1
Find the voltage drop across the 12 W resistance using Norton's for the circuit shown in Q 1.
[AU April/May 2011]
rrr 3-4.2 Explain in brief about source transform and Find Norton's equivalent circuit for the network shown
in Fig. Q.2 and obtain current in the 10 W resistor.
[GTU Dec. 2010]
4W
14 V
2W
5W
3W
4V
a
12 W
10V
4W
6W
1A
10W
b
Fig. Q.1
rrr 3-4.3
Fig. Q.2
In Fig. Q.3 if the 2 V source is replaced by an open circuit then find Thevenin's and Norton's
equivalent circuit across V2 and V3. Resistance values are in ohms.
[GTU May 2011]
127
Useful Theorems in Circuit Analysis
rrr 3-4.4
In the circuit shown in Fig. Q.4, find the current through RL connected across A-B terminals by
utilising Thevenin's theorem. Verify the results by Norton's theorem.
[JNTU Nov. 2012]
S
5W
A
4W
25 V
B
Fig. Q.3
rrr3-4.5
2W
5A
j5W
Fig. Q.4
Determine the current through the 1 W resistor if connected across AB in the network shown in Fig.
Q.5 using Norton's theorem.
[PU 2010]
Fig. Q.5
rrr 3-4.6
Obtain the Thevenin's and Norton's at the terminals A-B for the circuit shown in Fig. Q.6. [PU 2012]
2W
A
I
10 V
4W
5I
B
Fig. Q.6
3.5
RECIPROCITY THEOREM
In any linear bilateral network, if a single voltage source Va in branch ‘a’
LO 5
produces a current Ib in branch ‘b’, then if the voltage source Va is removed
and inserted in branch ‘b’ will produce a current Ib in branch ‘a’. The ratio of
response to excitation is same for the two conditions mentioned above. This is
called the reciprocity theorem.
Consider the network shown in Fig. 3.22. AA9 denotes input terminals and BB9 denotes output terminals.
Fig. 3.22
128
Circuits and Networks
The application of voltage V across AA9 produces current I at BB9. Now if the positions of the source and
responses are interchanged, by connecting the voltage source across BB9, the resultant current I will be at
terminals AA9. According to the reciprocity theorem, the ratio of response to excitation is the same in both
cases.
EXAMPLE 3.6
Verify the reciprocity theorem for the network shown in Fig. 3.23.
Fig. 3.23
Solution
Total resistance in the circuit 5 2 1 [3 || (2 1 2 || 2)] 5 3.5 V
The current drawn by the circuit (See Fig. 3.24 (a)).
20
IT =
= 5.71 V
3.5
The current in the 2 V branch cd is I 5 1.43 A.
Fig. 3.24
Applying the reciprocity theorem, by interchanging the source and response, we get Fig. 3.24 (b).
Fig. 3.24
Total resistance in the circuit 5 3.23 V.
20
Total current drawn by the circuit =
= 6.19 A
3.23
The current in the branch ab is I 5 1.43 A
If we compare the results in both cases, the ratio of input to response is the same, i.e. (20/1.43) 5 13.99
129
Useful Theorems in Circuit Analysis
Frequently Asked Questions linked to LO 5
rrr 3-5.1
rrr 3-5.2
What is the reciprocity theorem?
Verify the reciprocity theorem for the circuit shown below.
3W
100 V
[AU May/June 2014]
[AU Nov./Dec. 2012]
12 W
14 W
4W
4W
Fig. Q.2
3.6
COMPENSATION THEOREM
The compensation theorem states that any element in the linear, bilateral
network, may be replaced by a voltage source of magnitude equal to the
current passing through the element multiplied by the value of the element,
provided the currents and voltages in other parts of the circuit remain
unaltered. Consider the circuit shown in Fig. 3.25 (a). The element R can be
replaced by voltage source V, which is equal to the current I passing through
R multiplied by R as shown in Fig. 3.25 (b).
LO 6
Fig. 3.25
This theorem is useful in finding the changes in current or voltage when the value of resistance is changed
in the circuit. Consider the network containing a resistance R shown in Fig. 3.26 (a). A small change in
resistance R, that is (R 1 DR), as shown in Fig. 3.26 (b), causes a change in current in all branches. This
current increment in other branches is equal to the current produced by the voltage source of voltage I. DR
which is placed in series with altered resistance as shown in Fig. 3.26 (c).
Fig. 3.26
130
Circuits and Networks
EXAMPLE 3.7
Determine the current flowing in the ammeter having 1 V internal
resistance connected in series with a 3 V resistor as shown in
Fig. 3.27.
Fig. 3.27
Solution The current flowing through the 3 V branch is I3 5 1.11
Fig. 3.28
A. If we connect the ammeter having 1 V resistance to the 3 V
branch, there is a change in resistance. The changes in currents
in other branches then result as if a voltage source of voltage I3
DR 5 1.11 3 1 5 1.11 V is inserted in the 3 V branch as shown
in Fig. 3.28.
Current due to this 1.11 V source is calculated as follows.
Current I93 5 0.17 A
This current is opposite to the current I3 in the 3 V branch.
Hence the ammeter reading 5 (1.11 – 0.17) 5 0.94 A.
Frequently Asked Questions linked to LO 6
rr 3-6.1
In Fig. Q.1, if 1 ohm resistance is changed to 1.2 ohms
then determine the source-voltage for compensating for
the change.
[GTU May 2014]
I
2 V dc
V1
2
1 A dc
2
V2
2
4 V dc
GND
3.7
MAXIMUM POWER TRANSFER THEOREM
Fig. Q.1
Many circuits basically consist of sources, supplying voltage, current, or power
LO 7
to the load; for example, a radio speaker system, or a microphone supplying the
input signals to voltage pre-amplifiers. Sometimes it is necessary to transfer
maximum voltage, current or power from the source to the load. In the simple
resistive circuit shown in Fig. 3.29, Rs is the source resistance. Our aim is to
find the necessary conditions so that the power delivered by the source to the
load is maximum.
It is a fact that more voltage is delivered to the load when
the load resistance is high as compared to the resistance of the
source. On the other hand, maximum current is transferred to the
load when the load resistance is small compared to the source
resistance.
For many applications, an important consideration is the
maximum power transfer to the load; for example, maximum
power transfer is desirable from the output amplifier to the
Fig. 3.29
131
Useful Theorems in Circuit Analysis
speaker of an audio sound system. The maximum power transfer theorem states that maximum power is
delivered from a source to a load when the load resistance is equal to the source resistance. In Fig. 3.29,
assume that the load resistance is variable.
Current in the circuit is I 5 VS /(RS 1 RL)
Power delivered to the load RL is P 5 I 2RL 5 V S2 RL/(RS 1 RL)2
To determine the value of RL for maximum power to be transferred to the load, we have to set the first
dP
derivative of the above equation with respect to RL, i.e. when
equals zero.
dRL

VS2
dP
d 

R
=
L
dRL dRL  ( RS + RL )2 
=
{
VS2 ( RS + RL )2 − (2 RL ) ( RS + RL )
( RS + RL )
}
4
∴ (RS 1 RL)2 – 2RL(RS 1 RL) 5 0
R2S
1 R2L1 2RS RL – 2R2L – 2RS RL 5 0
∴ RS 5 RL
So, maximum power will be transferred to the load when load resistance is equal to the source resistance.
EXAMPLE 3.8
In the circuit shown in Fig. 3.30, determine the value of load
resistance when the load resistance draws maximum power.
Also find the value of the maximum power.
Solution In Fig. 3.30, the source delivers the maximum power
Fig. 3.30
when load resistance is equal to the source resistance.
RL 5 25 V
The current I 5 50/(25 1 RL) 5 50/50 5 1 A
The maximum power delivered to the load P 5 I 2RL
5 1 3 25 5 25 W
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 7
rr 3-7.1
For the circuit shown in Fig. Q.1, what will be
the value of RL to get the maximum power?
What is the maximum power delivered to the
load? What is the maximum voltage across
the load? What is the maximum current
in it?
Fig. Q.1
132
Circuits and Networks
For the circuit shown in Fig. Q.2, determine the value of RL to get the maximum power. Also find the
maximum power transferred to the load.
rrr 3-7.3 Using PSPICE, find the maximum power transferred to the resistor R in the following circuit
(Fig. Q.3).
rrr 3-7.2
10k
22k
40k
30k
Fig. Q.3
Fig. Q.2
Frequently Asked Questions linked to LO 7
rrr 3-7.1
rrr 3-7.2
State the maximum power transfer theorem for ac circuits.
[AU May/June 2013]
In the circuit of Fig. Q.2, find the value if R for maximum power transfer. Also calculate the
maximum power.
[AU May/June 2014]
10 W
15 W
12 V
2A
R
Fig. Q.2
rrr 3-7.3
Obtain Thevenin's equivalent circuit for the network shown Fig. Q.3, find the power dissipated in
RL = 5 W resistor. Find RL for maximum power transfer from the source and compute maximum
power that can be transferred, i.e. Pmaxi.
[GTU Dec. 2010]
a
10W
2A
20W
10 V
5W
b
Fig. Q.3
rrr 3-7.4
Prove the maximum power transfer theorem for a practical voltage source (VS). What is the
maximum power that can be delivered Vs = 20 V and Rs = 1 ohm?
[GTU May 2011]
rrr 3-7.5 Obtain the equivalent circuit A-B terminals in Fig. Q.5 and find the value of ZL to have maximum
power.
[JNTU Nov. 2012]
2W
j3 W
A
j1 W
100 V
j2 W
ZL
j4 W
B
Fig. Q.5
133
Useful Theorems in Circuit Analysis
rrr3-7.6
For the circuit shown in Fig. Q.6, find the value of R that will receive maximum power. Determine
this maximum power.
[PU 2012]
5.2 W
7.1 W
100 V
R
10.9 W
19.6 W
Fig. Q.6
3.8
DUALS AND DUALITY
In an electrical circuit itself there are pairs of terms which can be interchanged
LO 8
to get new circuits. Such pair of dual terms are given below.
Current — Voltage
Open — Short
L—C
R—G
Series — Parallel
Voltage source — Current source
KCL — KVL
Consider a network containing R-L-C elements connected in
series, and excited by a voltage source as shown in Fig. 3.31.
The integrodifferential equation for the above network is
Ri + L
Fig. 3.31
di 1
+
dt C
Similarly, consider a network containing R-L-C elements
connected in parallel and driven by a current source as shown in
Fig. 3.32.
The integrodifferential equation for the network in Fig. 3.32
is
i = Gv + C
Fig. 3.32
∫ idt = V
dv 1
+
vdt
dt L ∫
If we observe both the equations, the solutions of these two equations are the same. These two networks
are called duals.
To draw the dual of any network, the following steps are to be followed.
1. In each loop of a network place a node; and place an extra node, called the reference node, outside
the network.
2. Draw the lines connecting adjacent nodes passing through each element, and also to the reference
node, by placing the dual of each element in the line passing through original elements.
134
Circuits and Networks
For example, consider the network shown in Fig. 3.33.
Our first step is to place the nodes in each loop and a
reference node outside the network.
Drawing the lines connecting the nodes passing through
each element, and placing the dual of each element as
shown in Fig. 3.34 (a) we get a new circuit as shown in Fig.
3.34 (b).
Fig. 3.33
Fig. 3.34
EXAMPLE 3.9
Draw the dual network for the given network shown in Fig.
3.35.
Solution Place nodes in each loop and one reference node
outside the circuit. Joining the nodes through each element,
and placing the dual of each element in the line, we get the dual
circuit as shown in Fig. 3.36 (a).
Fig. 3.35
The dual circuit is redrawn as shown in Fig. 3.36 (b).
(a)
(b)
Fig. 3.36
135
Useful Theorems in Circuit Analysis
Frequently Asked Questions linked to LO 8
rrr 3-8.1
Explain dual with reference to network.
[GTU Dec. 2014]
Explain the principle of duality. Write a graphical procedure to draw a dual network?
[JNTU Nov. 2012]
rrr 3-8.3 Draw the dual of the network shown in Fig. Q.3 and explain its procedure.
[JNTU Nov. 2012]
rrr 3-8.2
Fig. Q.3
rrr 3-8.4
Explain the concept of duality. What relationship does duality have with the incidence matrix?
[PTU 2009-10]
rrr 3.8.5 Find the dual of the following network shown in Fig. Q.5.
[PTU 2009-10]
rrr 3.8.6 Draw the dual of the network shown in Fig. Q.6.
[RGTU Dec. 2013]
L1
L
L2
C
V (t)
R
Fig. Q.5
3.9
V (t)
C1
C2
C3
L3
R
L4
Fig. Q.6
TELLEGEN’S THEOREM
Tellegen’s theorem is valid for any lumped network which may be linear or
LO 9
nonlinear, passive or active, time-varying or time-invarient. This theorem states
that in an arbitrary lumped network, the algebraic sum of the powers in all
branches at any instant is zero. All branch currents and voltages in that network
must satisfy Kirchhoff’s laws. Otherwise, in a given network, the algebraic sum
of the powers delivered by all sources is equal to the algebraic sum of the powers absorbed by all elements.
This theorem is based on Kirchhoff’s two laws, but not on the type of circuit elements.
Consider two networks N1 and N2, having the same graph with different types of elements between the
corresponding nodes.
b
Then
∑ v1K i2 K = 0
K =1
b
and
∑ v2 K i1K = 0
K =1
To verify Tellegen’s theorem, consider two circuits having same graphs as shown in Fig. 3.37.
136
Circuits and Networks
In Fig. 3.37 (a),
i1 5 i2 5 2 A; i3 5 2 A
and v1 5 –2 V, v2 5 –8 V, v3 5 10 V
In Fig. 3.37 (b),
Fig. 3.37 (a)
i11 = i21 = 4 A; i31 = 4 A
and v11 = −20 V; v12 = 0 V; v31 = 20 V
3
Now
∑ vK i1K = v1 i11 + v2 i21 + v3 i31
K =1
= ( − 2) (4) + ( − 8) (4) + (10) (4) = 0
3
Fig. 3.37 (b)
and
∑ v1K iK = v11 i1 + v12 i2 + v31 i3
K =1
= ( − 20) (2) + (0) (2) + (20) (2) = 0
Similarly,
3
∑ vK iK = v1i1 + v2i2 + v3i3
K =1
= (–2) (2) + (–8) (2) + (10) (2) = 0
3
and
∑ v1K i1K = (−20)(4) + (0)(4) + (20)(4) = 0
K =1
This verifies Tellegen’s theorem.
Frequently Asked Questions linked to LO 9
rrr3-9.1
rrr 3-9.2
Fig. Q.1
3.10
[PTU 2009-10]
[PTU 2011-12]
Verify Tellegen's theorem for the network shown in Fig. Q.1.
Verify Tellegen's theorem for the pair of networks.
Fig. Q.2
MILLMAN’S THEOREM
Millman’s theorem states that in any network, if the voltage sources V1, V2, …,
Vn in series with internal resistances R1, R2, …, Rn, respectively, are in parallel,
then these sources may be replaced by a single voltage source V9in series with
R9 as shown in Fig. 3.38.
LO 10
137
Useful Theorems in Circuit Analysis
Fig. 3.38
where V ′ =
V1G1 + V2G2 + �VnGn
G1 + G2 + � + Gn
Here, Gn is the conductance of the nth branch,
1
and R ′ =
G1 + G2 + � + Gn
A similar theorem can be stated for n current sources having internal conductances which can be replaced
by a single current source I9 in parallel with an equivalent conductance.
where
I′=
and G ′ =
I1 R1 + I 2 R2 + � I n Rn
R1 + R2 + � + Rn
1
R1 + R2 + � + Rn
Fig. 3.39
EXAMPLE 3.10
Calculate the current I shown in Fig. 3.40 using
Millman’s Theorem.
Solution According to Millman’s theorem, the two
voltage sources can be replaced by a single voltage
source in series with resistance as shown in Fig. 3.41.
We have V 9 =
=
and R9=
Fig. 3.40
V1G1 + V2G2
G1 + G2
[10(1 2) + 20(1 5) ]
1 2 +1 5
= 12.86 V
1
1
=
=1.43 V
G1 + G2 1 / 2 +1 / 5
Fig. 3.41
138
Circuits and Networks
Therefore, the current passing through the 3 V resistor is
I=
12.86
= 2.9 A
3 +1.43
Frequently Asked Questions linked to LO 10
rr 3-10.1 State Millman's theorem. Obtain the equivalent of a parallel connection of three branches each with
a voltage source and a series resistance (2 V, 1 ohm), (3 V, 2 ohms), (5 V, 2 ohms). [GTU May 2011]
rr 3-10.2 State and explain the following.
(a) Reciprocity theorem
(b) Millman's theorem
(RGTU Dec. 2013)
Additional Solved Problems
PROBLEM 3.1
Calculate the voltage across AB in the network shown in Fig.
3.42 and indicate the polarity of the voltage using star-delta
transformation.
Solution The circuit in Fig. 3.42 can be redrawn as shown
Fig. 3.42
in Fig. 3.43.
By taking star-delta transformation at ABC, we have
The circuit shown in Fig. 3.44 is a single node circuit and V is
the voltage at mode.
Applying nodal analysis, we have
V
V
V −10
+
+
=0
4.31 6.77
2.92
From Eq. (3.8), the voltage is V 5 4.744 volts
Fig. 3.43
The voltage across AB is VAB5 VA– VB
2
2
VA = V ×
= 4.744 ×
2 + 2.31
2 + 2.31
= 2.2 volts
6
6
= 4.744 ×
6 + 0.77
6.77
= 4.2 volts
VB = V ×
∴
Fig. 3.44
the voltage across AB is VAB 5 VA – VB
VAB 5 2.2 – 4.2 5 – 2 volts
(3.8)
Useful Theorems in Circuit Analysis
139
PROBLEM 3.2
Determine the current I in the circuit shown in Fig. 3.45 using the superposition theorem.
Fig. 3.45
Solution
The current I' due to 10 V source, when other sources are zero, is shown in Fig. 3.46.
From the Fig. 3.46 (a), the resistances of 2 V, 3 V, and 10 V are in parallel and resultant is in series with 5 V
and 3 V. The equivalent circuit is shown in Fig. 3.46 (b).
Fig. 3.46
The total equivalent resistance in Fig. 3.46 (b) is
Req 5 [(8 1 1.07) //1] 1 1 5 1.9 V
The current passing through the 1 V resistance is
I′=
10
= 5.26 A
1.9
The current I'' due to the 5 V source, when other sources are zero, is shown in Fig. 3.47.
The circuit in Fig. 3.47 (a) can be further reduced as shown in Fig. 3.47 (b).
Total current delivered by the 5 V source
IT =
5
= 0.86 A
5.8
140
Circuits and Networks
The current passing through the 5 V resistor
Fig. 3.47
0.86 ×10
= 0.46 A
18.5
The current passing through the 1 V resistor
I5 =
0.46 ×1
= 0.23 A
2
The current I''' due to the 10 A source, when other sources are set to zero as shown in Fig. 3.48.
I" =
Fig. 3.48
The circuit in Fig. 3.48 (a) can be further simplified as shown in Fig. 3.48 (b).
The current passing through the 3 V resistance =
The current passing through the 1 V resistor is
1
I ′′′ = 6.34 × = 3.17 A
2
The current passing through the 1 V resistor is
I = I ′ − I ′′ − I ′′′
= 5.26 − 0.23 − 3.17
= 1.86 A
10 ×6.07
= 6.34 A
6.07 + 3.5
141
Useful Theorems in Circuit Analysis
PROBLEM 3.3
Find the Thevenin’s and Norton’s equivalents
for the circuit shown in Fig. 3.49 with respect to
terminals ab.
Solution To find the Thevenin’s resistance,
Fig. 3.49
voltage sources are to be short circuited.
Therefore, the resistance seen into the terminals ab is
Rab 5 [20V||40V] 5 13.33 V
RTh 5 RN 5 Rab 5 13.33 V
In the circuit shown in Fig. 3.50 (a), the open circuit voltage across terminals
ab 5 Vab 5 10 1 I20
5 10 1 I(20)
where
I=
50 −10 40
=
= 0.67A
40 + 20 60
∴ Thevenin’s voltage Vab 5 10 1 (0.67)20 5 23.4 1 V
The short-circuit current through terminals ab as shown in Fig. 3.50 (b) is
IN 5 I1 1 I2
where
∴
I1 =
10
50
= 0.5 A and I 2 =
= 1.25 A
20
40
IN 5 1.75 A
∴ Norton’s current IN 5 1.75 A
Fig. 3.50
The Thevenin’s equivalent circuit is shown in Fig. 3.51 (a) and the Norton’s equivalent circuit is shown in
Fig. 3.51 (b).
142
Circuits and Networks
Fig. 3.51
PROBLEM 3.4
Determine the Thevenin’s and Norton’s equivalent circuits with respect to terminals ab for the circuit shown
in Fig. 3.52.
a
b
Fig. 3.52
Solution To find out the resistance seen into the terminals, when all the current sources are open circuited.
Rab 5 4 V
By applying the superposition theorem, we get the voltage across the 4 V resistor is
Vab 5 V4 5 V12 1 V5 1 V3
The voltage across the 4 V resistor due to the 12 A source when other sources are set equal to zero is
V12 5 0
The voltage across the 4 V resistor due to the 5 A source when other sources are set equal to zero is
V5 5 5 3 4 5 20 V
The voltage across 4 V resistor due to 3 A source when other sources are set equal to zero
V3 5 4 3 3 5 12 V
∴ Vab 5 0 1 20 1 12 5 32 V
The current in the short-circuited terminals
Iab 5 I12 1 I5 1 I3
The current due to only the 12 A source is I12 5 0 A
143
Useful Theorems in Circuit Analysis
The current due to only the 5 A source is I5 5 5 A
The current due to only the 3 A source is I3 5 3 A
Iab 5 0 1 5 1 3 5 8 A
Therefore, the Thevenin’s equivalent circuit is shown in Fig. 3.53 (a) and the Norton’s equivalent circuit is
shown in Fig. 3.53 (b).
Fig. 3.53
PROBLEM 3.5
By using source transformation or any other technique, replace the circuit shown in Fig. 3.54 between
terminals ab with the voltage source in series with a single resistor.
Fig. 3.54
Solution From Fig. 3.54, the resistances of 30 V and 50 V are in parallel and the resultant resistance (30 V
||50 V) 5 18.75 V in parallel with the 3 A current source can be replaced by an equivalent voltage source in
series with the resistance as shown in Fig. 3.55 (b).
a
a
b
b
Fig. 3.55
144
Circuits and Networks
Considering the node voltage Vab, by applying Kirchhoff’s current law, we have
Vab − 56.25 Vab − 20 Vab
+
+
=0
5
6
18.75
Vab [0.05 1 0.2 1 0.17]–3– 4 5 0
∴ Thevenin’s voltage Vab 5 16.67 V
Resistance seen into the terminals ab when the voltage sources are short circuited in Fig. 3.55 (b).
Thevenin’s resistance Rab 5 [18.75||5||6] 5 2.38 V
∴ the Thevenin’s equivalent circuit is shown in Fig. 3.56.
Fig. 3.56
PROBLEM 3.6
Use Thevenin’s theorem to find the current through the 5 V
resistor in Fig. 3.57.
Solution Thevenin’s equivalent circuit can be formed by
obtaining the voltage across terminals AB as shown in Fig. 3.58
(a).
Current in the 6 V resistor, I 6 =
100
= 6.25 A
16
Voltage across the 6 V resistor, V6 = 6 ×6.25 = 37.5 V
100
= 4.35 A
23
Voltage across the 8 V resistor is V8 5 4.35 3 8 5 34.8 V
Current in the 8 V resistor, I 8 =
Voltage across the terminals AB is VAB 5 37.5 – 34.8 5 2.7 V
The resistance as seen into the terminals RAB
=
6 ×10 8×15
+
6 + 10 8 + 15
5 3.75 1 5.22 5 8.97 V
Thevenin’s equivalent circuit is shown in Fig. 3.58 (b).
2.7
Current in the 5 V resistor, I 5 =
= 0.193 A
5 + 8.97
Fig. 3.57
145
Useful Theorems in Circuit Analysis
Fig. 3.58
PROBLEM 3.7
Find Thevenin’s equivalent circuit for the circuit shown in
Fig. 3.59.
Solution Thevenin’s voltage is equal to the voltage across the
terminals AB.
Fig. 3.59
VAB 5 V3 1 V6 1 10
Here, the current passing through the 3 V resistor is zero.
Hence, V3 5 0
By applying Kirchhoff’s law, we have
50 −10 = 10 I + 6 I
40
= 2.5 A
16
The voltage across 6 V is V6 with polarity as shown in Fig. 3.60 (a), and is given by
I=
V6 5 6 3 2.5 5 15 V
The voltage across terminals AB is VAB 5 0 1 15 1 10 5 25 V.
The resistance as seen into the terminals AB
10×6
= 6.75 V
10 + 6
Thevenin’s equivalent circuit is shown in Fig. 3.60 (b).
RAB = 3 +
Fig. 3.60
146
Circuits and Networks
PROBLEM 3.8
Determine the Thevenin’s equivalent circuit across terminals AB for the circuit in Fig. 3.61.
Fig. 3.61
Solution The given circuit is redrawn as shown in Fig. 3.62 (a).
Voltage VAB 5 V2 1 V1
Fig. 3.62
Applying Kirchhoff’s voltage law to loops 1 and 2, we have the following:
Voltage across the 2 V resistor
Voltage across the 1 V resistor
10
= 2.85 V
7
5
V1 = 1× = 1 V
5
V2 = 2×
∴ VAB 5 V2 1 V1
5 2.85 – 1 5 1.85 V
The resistance seen into the terminals AB
RAB 5 (5 || 2) 1 (4 || 1)
=
Fig. 3.62
5× 2 4 ×1
+
5 + 2 4 +1
5 1.43 1 0.8 5 2.23 V
Thevenin’s equivalent circuit is shown in Fig. 3.62 (b).
147
Useful Theorems in Circuit Analysis
PROBLEM 3.9
Determine Norton’s equivalent circuit for the circuit shown in
Fig. 3.63.
Fig. 3.63
Solution Norton’s equivalent circuit is given by Fig. 3.64 (a)
where IN 5 short-circuit current at terminals AB
RN 5 open-circuit resistance at terminals AB
The current IN can be found as shown in Fig. 3.64 (b).
Fig. 3.64
IN =
50
= 16.7 A
3
Norton’s resistance can be found from Fig. 3.64 (c).
RN = RAB =
3× 4
= 1.71 V
3+ 4
Norton’s equivalent circuit for the given circuit is shown in Fig. 3.64 (d).
Fig. 3.64
148
Circuits and Networks
PROBLEM 3.10
Determine Norton’s equivalent circuit for the given
circuit shown in Fig. 3.65.
Fig. 3.65
The short-circuit current at terminals AB can be found from Fig. 3.66 (a) and Norton’s resistance
can be found from Fig. 3.66 (b).
Solution
The current IN is same as the current in the 3 V resistor or 4 V resistor.
I N = I 3 = 25×
2
= 5.55 A
7+2
The resistance as seen into the terminals AB is
RAB 5 5 || (4 1 3 1 2)
=
Fig. 3.66
5× 9
= 3.21 V
5+9
Norton’s equivalent circuit is shown in Fig. 3.66 (c).
PROBLEM 3.11
Determine the current flowing through the 5 V resistor in the circuit shown in Fig. 3.67 by using Norton’s
theorem.
149
Useful Theorems in Circuit Analysis
Fig. 3.67
Solution The short-circuit current at terminals AB can be found from the circuit as shown in Fig. 3.68 (a).
Norton’s resistance can be found from Fig. 3.63 (b).
In Fig. 3.68 (a), the current IN 5 30 A.
Fig. 3.68
The resistance in Fig. 3.68 (b)

1×1
RAB = 5 � 2 +


2 
5× 2.5
= 5 � ( 2.5) =
= 1.67 V
7.5
Norton’s equivalent circuit is shown in Fig. 3.68 (c).
∴ the current in the 5 V resistor
Fig. 3.68
I 5 = 30 ×
1.67
= 7.51 A
6.67
PROBLEM 3.12
Replace the given network shown in Fig. 3.69 by a single current
source in parallel with a resistance.
Fig. 3.69
Solution Here, using superposition technique and Norton’s theorem, we can convert the given network.
We have to find a short-circuit current at terminals AB in Fig. 3.70 (a) as shown.
The current I9N is due to the 10 A source. I9N 5 10 A
The current I99N is due to the 20 V source (see Fig. 3.70 (b) and (c)).
150
Circuits and Networks
Fig. 3.70
20
= 3.33 A
6
The current IN is due to both the sources
I N′′ =
IN 5 I9N 1 IN
5 10 1 3.33 5 13.33 A
The resistance as seen from terminals AB
RAB 5 6 V (from Fig. 3.70 (d))
Hence, the required circuit is as shown in Fig. 3.70 (e).
Fig. 3.70
Fig. 3.70
PROBLEM 3.13
Using the compensation theorem, determine the ammeter
reading where it is connected to the 6 V resistor as shown
in Fig. 3.71. The internal resistance of the ammeter is 2 V.
Solution The current flowing through the 5 V branch
Fig. 3.71
3
I 5 = 20 ×
= 6.315 A
3 + 6.5
So the current in the 6 V branch
I 6 = 6.315×
2
= 1.58 A
6+2
If we connect the ammeter having 2 V internal resistance to the 6 V branch, there is a change in resistance. The
changes in currents in other branches results if a voltage source of voltage I6. DR 5 1.58 3 2 5 3.16 V is
inserted in the 6 V branch as shown in Fig. 3.72.
151
Useful Theorems in Circuit Analysis
The current due to this 3.16 V source is calculated.
The total impedance in the circuit
RT 5 {[(6 || 3) 1 5] || [2]} 1 {6 1 2}
5 9.56 V
The current due to the 3.16 V source
3.16
I 6′ =
= 0.33 A
9.56
This current is opposite to the current I6 in the 6 V branch.
Hence, the ammeter reading 5 (1.58 – 0.33)
Fig. 3.72
5 1.25 A
PROBLEM 3.14
Verify the reciprocity theorem for the given circuit shown
in Fig. 3.73.
In Fig. 3.73, the current in the 5 V resistor is
4
4
I5 = I 2 ×
= 2.14 ×
8+4
12
= 0.71 A
10
where I 2 =
R
Solution
Fig. 3.73
RT = 4.67
10
∴ I2 =
= 2.14 A
4 67
and
We interchange the source and response as shown in Fig. 3.74.
In Fig. 3.74, the current in the 2 V resistor is
I 2 = I3 ×
where
and
Fig. 3.74
I3 =
4
4+2
10
RT
RT = 9.33 V
∴ I3 =
10
= 1.07 A
9.33
4
I 2 = 1.07× = 0.71 A
6
In both cases, the ratio of voltage to current is
Hence, the reciprocity theorem is verified.
10
= 14.08.
0.71
152
Circuits and Networks
PROBLEM 3.15
Verify the reciprocity theorem in the circuit shown in Fig.
3.75.
Solution The voltage V across the 3 V resistor is
where
V 5 I3 3 R
2
I 3 = 10 ×
= 4A
2+3
Fig. 3.75
V 5 4 3 3 5 12 V
We interchange the current source and response as shown in Fig. 3.76.
To find the response, we have to find the voltage across the
2 V resistor
where
V 5 I2 3 R
3
I 2 = 10 × = 6 A
5
V 5 6 3 2 5 12 V
Fig. 3.76
In both cases, the ratio of current to voltage is the same, i.e.
it is equal to 0.833. Hence, the reciprocity theorem is verified.
PROBLEM 3.16
Determine the maximum power delivered to the load in the
circuit shown in Fig. 3.77.
Solution For the given circuit, let us find out the Thevenin’s
equivalent circuit across AB as shown in Fig. 3.78 (a).
The total resistance is
RT 5 [{(3 1 2) || 5} 1 10]
5 [2.5 1 10] 5 12.5 V
Total current drawn by the circuit is
IT =
50
= 4A
12.5
The current in the 3 V resistor is
I 3 = IT ×
5
4 ×5
=
= 2A
5+5
10
Fig. 3.77
153
Useful Theorems in Circuit Analysis
Thevenin’s voltage VAB 5 V3 5 3 3 2 5 6 V
Thevenin’s resistance RTh 5 RAB 5 [((10 || 5) 1 2) || 3] V 5 1.92 V
Thevenin’s equivalent circuit is shown in Fig. 3.78 (b).
Fig. 3.78
From Fig. 3.78 (b) and maximum power transfer theorem,
RL 5 1.92 V
∴ current drawn by the load resistance RL
6
IL =
= 1.56 A
1.92 + 1.92
Power delivered to the load 5 IL2 RL
5 (1.56)2 3 1.92 5 4.67 W
PROBLEM 3.17
Determine the load resistance to receive maximum power from
the source; also find the maximum power delivered to the load
in the circuit shown in Fig. 3.79.
Solution For the given circuit, we find out the Thevenin’s
equivalent circuit.
Thevenin’s voltage across terminals A and B
Fig. 3.79
VAB 5 VA – VB
30
= 75 V
30 + 10
40
Voltage at the point B is VB = 100 ×
= 66.67 V
40 + 20
Voltage at the point A is VA = 100 ×
VAB 5 75 – 66.67 5 8.33 V
To find Thevenin’s resistance, the circuit in Fig. 3.80
(a) can be redrawn as shown in Fig. 3.80 (b).
Fig. 3.80
154
Circuits and Networks
(b)
Fig. 3.80
From Fig. 3.80 (b), Thevenin’s resistance
RAB 5 [(30 || 10) 1 (20 || 40)]
5 [7.5 1 13.33] 5 20.83 V
Thevenin’s equivalent circuit is shown in Fig. 3.80 (c).
According to maximum power transfer theorem,
RL 5 20.83 V
Current drawn by the load resistance
IL =
8.33
= 0.2 A
20.83 + 20.83
Fig. 3.80
∴ maximum power delivered to load 5 I L2 RL
5 (0.2)2 (20.83) 5 0.833 W
PROBLEM 3.18
Draw the dual circuit for the given circuit shown in Fig. 3.81.
Our first step is to place nodes in each loop, and a reference
node outside the circuit.
Solution
Join the nodes with lines passing through each element and connect
these lines with the dual of each element as shown in Fig. 3.82 (a).
(a)
Fig. 3.82
Fig. 3.81
155
Useful Theorems in Circuit Analysis
The dual circuit of the given circuit is shown in Fig. 3.82 (b).
(b)
Fig. 3.82
PROBLEM 3.19
Draw the dual circuit of Fig. 3.83 given below.
Solution Our first step is to mark nodes in each loop and a
reference node outside the circuit.
Join the nodes with lines passing through each element and
connect these lines with the dual of each element as shown in Fig.
3.84 (a).
The dual circuit of the given circuit is shown in Fig. 3.84 (b).
Fig. 3.84
PROBLEM 3.20
For the circuit shown in Fig. 3.85, find the current i4 using the superposition principle.
Fig. 3.83
156
Circuits and Networks
Fig. 3.85
Solution The circuit can be redrawn as shown in Fig. 3.86 (a).
The current i94 due to the 20 V source can be found using the circuit shown in Fig. 3.86 (b).
Fig. 3.86
Applying Kirchhoff’s voltage law,
– 20 1 4i94 1 2i94 1 2i94 5 0
i94 5 2.5 A
The current i''4 due to the 5 A source can be found using the circuit shown in Fig. 3.86 (c).
By assuming V'' at the node shown in Fig. 3.86 (c) and applying Kirchhoff’s current law,
V ′′ − 2i ′′4
V ′′
−5 +
=0
4
2
−V ′′
i ′′4 =
4
From the above equations,
i''4 5 – 1.25 A
Fig. 3.86
∴ total current i4 5 i''4 1 i''4 5 1.25 A
PROBLEM 3.21
Determine the current through the 2 V resistor as shown in
Fig. 3.87 by using the superposition theorem.
Solution The current I9 due to the 5 V source can be found using
the circuit shown in Fig. 3.88 (a).
By applying Kirchhoff’s voltage law, we have
3I9 1 5 1 2I9 – 4V93 5 0
Fig. 3.87
157
Useful Theorems in Circuit Analysis
we know
V93 5 – 3I9
From the above equations,
I9 5 – 0.294 A
The current I'' due to the 4 A source can be found using the circuit shown in Fig. 3.88 (b).
Fig. 3.88
By assuming the node voltage V''3, we find
I" =
V3" + 4V3"
2
By applying Kirchhoff’s current law at the node, we have
V3"
V " + 4V3"
−4+ 3
=0
3
2
V''3 5 1.55 V
V3" + 4V3"
= 3.875 A
2
Total current in the 2 V resistor I 5 I9 1 I'' 5 – 0.294 1 3.875
∴
I" =
∴ I 5 3.581 A
PROBLEM 3.22
For the circuit shown in Fig. 3.89, obtain Thevenin’s equivalent
circuit.
Solution The circuit consists of a dependent source. In the presence of
the dependent source, RTh can be determined by finding vOC and iSC
∴
RTh =
vOC
iSC
Open-circuit voltage can be found from the circuit shown in Fig. 3.90 (a).
Since the output terminals are open, current passes through the 2 V branch only.
vx 5 2 3 0.1 vx 1 4
4
vx =
= 5V
0.8
Fig. 3.89
158
Circuits and Networks
Short-circuit current can be calculated from the circuit shown in
Fig. 3.90 (b).
Since vx 5 0, the dependent current source is opened.
4
The current iSC =
= 0.8 A
2+3
∴
Fig. 3.90
RTh =
vOC
5
=
= 6.25 V
iSC
0.8
The Thevenin’s equivalent circuit is shown in Fig. 3.90 (c).
PROBLEM 3.23
For the circuit shown in Fig. 3.91, find the current i2 in the 2 V
resistor by using Thevenin’s theorem.
Solution From the circuit, there is open voltage at the terminals
ab which is
Fig. 3.91
VOC 5 – 4Vi
where Vi 5 –4Vi – 5
∴ Vi 5 –1
Thevenin’s voltage VOC 5 4 V
From the circuit, short-circuit current is determined by shorting
terminals a and b.
Applying Kirchhoff’s voltage law, we have
4Vi 1 2iSC 5 0
Fig. 3.92
∴
RTh =
We know Vi 5 – 5
Substituting Vi in the above equation, we get
iSC 5 10 A
VOC
4
= = 0.4 V
10
iSC
The Thevenin’s equivalent circuit is as shown in Fig. 3.92.
4
= 1.67 A
The current in the 2 V resistor i2 =
2.4
159
Useful Theorems in Circuit Analysis
PROBLEM 3.24
For the circuit shown in Fig. 3.93, find Norton’s equivalent circuit.
Fig. 3.93
Solution In the case of circuit having only dependent sources (without independent sources), both VOC and iSC
V
1
or we can also apply the 1 V source externally and determine the
current through it and then we find RTh 5 1/i.
are zero. We apply a 1 A source externally and determine the resultant voltage across it, and then find RTh =
By applying the 1 A source externally as shown in Fig. 3.94
(a) and application of Kirchhoff’s current law, we have
Vx Vx + 4Vx
+
=1
5
2
Vx 5 0.37 V
The current in the 4 V branch is
Vx −V
= −1
4
Substituting Vx in the above equation, we get
V 5 4.37 V
Fig. 3.94
∴
RTh =
V
= 4.37 V
1
If we short circuit the terminals a and b, we have
Vx − 4Vx
=0
2
Vx 5 0
I SC =
Vx
=0
4
Therefore, Norton’s equivalent circuit is as shown in Fig. 3.94 (b).
160
Circuits and Networks
PSpice Problems
PROBLEM 3.1
Determine the voltage across the terminal AB of the circuit shown in
Fig. 3.95.
* TO DETERMINE THEVENIN VOLTAGE
V1 1
2
6
I1
0
2
5
R1 1
2
5
V2 2
0
10 V
.TF V(1,0) V1 ; TRANSFER FUNCTION ANALYSIS
.END
**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE 5 27.000 DEG C
****************************************************************
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
(1) 16.0000 (2) 10.0000
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V1
–1.200E 1 00
V2 5.000E 1 00
**** SMALL-SIGNAL CHARACTERISTICS
V(1,0)/ V1 5 1.000E 1 00
INPUT RESISTANCE AT V1 5 5.000E 1 00
OUTPUT RESISTANCE AT V(1,0) 5 0.000E 1 00
Fig. 3.95
Result
VAB 5 V(1,0) 5 16 V.
V(1,0) 5 1 3 V1 5 16 V 5 Vth.
PROBLEM 3.2
Use Thevenin's theorem to find the current through the 5 V
resistor in Fig. 3.96.
* TO DETERMINE THEVENIN CIRCUIT
VS 10
DC
100
R1 1
2
10
R2 2
0
6
R3 1
3
15
R4 3
0
8
.TF V(2, 3) VS
.END
Fig. 3.96
Useful Theorems in Circuit Analysis
OUTPUT
**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE 5 27.000 DEG C
****************************************************************
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
(1) 100.0000 (2) 37.5000
(3) 34.7830
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
VS
–1.060E 1 01
**** SMALL-SIGNAL CHARACTERISTICS
V(2,3)/VS 5 2.717E – 02
INPUT RESISTANCE AT VS 5 9.436E 1 00
OUTPUT RESISTANCE AT V(2,3) 5 8.967E 1 00
Result
VTH 5 V(2,3)/VS*VS 5 2.717E – 02 * 100 5 2.717 V
RH 5 OUTPUT RESISTANCE AT V(2,3) 5 8.967 V.
RL 5 5 V ; IL 5 VTH/(RTH 1 RL) 5 2.717/(8.967 1 5) 5 0.195A.
PROBLEM 3.3
Determine the load resistance to receive maximum power from
the source; also find the maximum power delivered to the load
in the circuit shown in Fig. 3.97 using PSpice.
* NETLIST TO FIND MAX.POWER TRANSFER
VS 10
DC
100
R1 1
2
10
R2 2
0
30
Fig. 3.97
R3 1
3
20
R4 3
0
40
RL 23 RLOAD 1
.MODEL RLOAD RES(R 5 25)
.DC RES RLOAD(R) 0.001 40 0.01
.TF V(2,3) VS
.PROBE
.END
OUTPUT
**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE 5 27.000 DEG C
****************************************************************
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
(1) 100.0000
(2) 73.6360
(3) 69.0910
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
VS
– 4.182E 1 00
TOTAL POWER DISSIPATION 4.18E 1 02 WATTS
161
162
Circuits and Networks
**** SMALL-SIGNAL CHARACTERISTICS
V(2,3)/ VS 5 4.545E – 02
INPUT RESISTANCE AT VS 5 2.391E 1 01
OUTPUT RESISTANCE AT V(2,3) 5 1.136E 1 01
Result
Fig. 3.98
From the graph shown in Fig. 3.98, which is obtained with .PROBE statement for W(RL), MAX(W(RL)) is
found to be 0.833W at RL 5 20.81 V.
Useful Theorems in Circuit Analysis
163
Answers to Practice Problems
3-1.1 1.182 V
3-1.2 Req 5 0.6078
3-1.3 (i) I6.8k 5 2.55 mA; I2.2k 5 7.9 mA
(ii) V6.8k 5 17.34 V ; V2.2k 5 17.38 V
3-2.1 0.82 A
3-2.2 I4V 5 1.5 V
3-2.3 I1 5 4.6 A; I2 5 2.6 A; I3 5 2 A
3-2.4 4 A
3-3.1 0.5 A
3-4.1 (i) 8 V voltage source is the series with 10 kV resistance
(ii) 0.8 mA current source is in parallel with 10 kV resistance
3-4.2
3-7.1
3-10.1
12 V, 0.75 W, 6 V, 0.5 A
I10 5 0.155 A
Objective-Type Questions
r 3.1
Three equal resistances of 3 V are connected in star. What is the resistance in one of the arms in an equivalent
delta circuit?
(a) 10 V
(b) 3 V
(c) 9 V
(d) 27 V
r 3.2 Three equal resistances of 5 V are connected in delta. What is the resistance in one of the arms of the
equivalent star circuit?
(a) 5 V
(b) 1.33 V
(c) 15 V
(d) 10 V
r 3.3 Superposition theorem is valid only for
(a) linear circuits
(c) both linear and nonlinear
(b) nonlinear circuits
(d) neither of the two
r 3.4 Superposition theorem is not valid for
(a) voltage responses
(c) power responses
(b) current responses
(d) all the three
Fig. 3.99
rr 3.5 Determine the current I in the circuit shown in Fig. 3.99. It is
(a) 2.5 A
(b) 1 A
(c) 3.5 A
(d) 4.5 A
r 3.6 Reduce the circuit shown in Fig. 3.100 to its Thevenin equivalent
circuit as viewed from terminals A and B.
(a) The circuit consists of 15 V battery in series with 100 kV
(b) The circuit consists of 15 V battery in series with 22 kV
(c) The circuit consists of 15 V battery in series with parallel
combination of 100 kV and 22 kV
(d) None of the above
Fig. 3.100
164
rrr 3.7
rrr 3.8
rrr 3.9
rrr 3.10
rrr 3.11
rrr 3.12
rrr 3.13
rrr 3.14
rrr 3.15
rrr 3.16
Circuits and Networks
Norton’s equivalent circuit consists of
(a) voltage source in parallel with resistance
(b) voltage source in series with resistance
(c) current source in series with resistance
(d) current source in parallel with resistance
The reciprocity theorem is applicable to
(a) linear networks only
(c) linear/bilateral networks
(b) bilateral networks only
(d) neither of the two
Compensation theorem is applicable to
(a) linear networks only
(c) linear and nonlinear networks
(b) nonlinear networks only
(d) neither of the two
Maximum power is transferred when load impedance is
(a) equal to source resistance
(c) equal to zero
(b) equal to half of the source resistance
(d) none of the above
In the circuit shown in Fig. 3.101, what is the maximum
power transferred to the load?
(a) 5 W
(c) 10 W
(b) 2.5 W
(d) 25 W
Indicate the dual of a series network that consists of voltage
Fig. 3.101
source, capacitance, inductance in
(a) parallel combination of resistance, capacitance, and inductance
(b) series combination of current source, capacitance, and inductance
(c) parallel combination of current source, inductance, and capacitance
(d) none of the above
When the superposition theorem is applied to any circuit, the dependent voltage source in that circuit is always.
(a) opened
(b) shorted
(c) active
(d) none of the above
Superposition theorem is not applicable to networks containing.
(a) nonlinear elements
(c) dependent current sources
(b) dependent voltage sources
(d) transformers
Thevenin's voltage in the circuit shown in Fig. 3.102 is
(a) 3 V
(b) 2 V
(c) 2.5 V
(d) 0.1 V
Norton’s current in the circuit shown in Fig. 3.103 is
(a)
2i
5
(b) infinite
(c) zero
(d) none
rrr 3.17 A dc circuit shown in Fig. 3.104 has a voltage V, a current source I and several resistors. A particular resistor R
dissipates a power of 4 W when V alone is active. The same resistor dissipates a power of 9 W when I alone is
active. The power dissipated by R when both sources are active will be
(a) 1 W
(b) 13 W
(c) 5 W
(d) 25 W
Resistive
network
Fig. 3.102
Fig. 3.103
For interactive quiz with answers,
scan the QR code given here
OR
visit
http://qrcode.flipick.com/index.php/261
Fig. 3.104
4
LEARNING OBJECTIVES
4.1
THE SINE WAVE
Many a time, alternating voltages and currents are represented by a sinusoidal
LO 1
wave, or simply a sinusoid. It is a very common type of alternating current (ac) and
alternating voltage. The sinusoidal wave is generally referred to as a sine wave.
Basically, an alternating voltage (current) waveform is defined as the voltage
(current) that fluctuates with time periodically, with change in polarity and direction. In general, the sine
wave is more useful than other waveforms, like pulse, sawtooth, square, etc. There are a number of reasons
for this. One of the reasons is that if we take any second- order system, the response of this system is
a sinusoid. Secondly, any periodic waveform can be written
in terms of sinusoidal function according to Fourier theorem.
Another reason is that its derivatives and integrals are also
sinusoids. A sinusoidal function is easy to analyse. Lastly, the
sinusoidal function is easy to generate, and it is more useful
in the power industry. The shape of a sinusoidal waveform is
shown in Fig. 4.1.
The waveform may be either a current waveform, or a
voltage waveform. As seen from Fig. 4.1, the wave changes
its magnitude and direction with time. If we start at time
Fig. 4.1
t 5 0, the wave goes to a maximum value and returns to zero,
and then decreases to a negative maximum value before returning to zero. The sine wave changes with
time in an orderly manner. During the positive portion of voltage, the current flows in one direction; and
during the negative portion of voltage, the current flows in the opposite direction. The complete positive
and negative portion of the wave is one cycle of the sine wave. Time is designated by t. The time taken
166
Circuits and Networks
Fig. 4.2
for any wave to complete one full cycle is called the period
(T ). In general, any periodic wave constitutes a number of
such cycles. For example, one cycle of a sine wave repeats
a number of times as shown in Fig. 4.2. Mathematically, it
can be represented as f(t) 5 f(t 1 T ) for any t.
The period can be measured in the following different
ways (see Fig. 4.3).
1. From zero crossing of one cycle to zero crossing of
the next cycle
2. From positive peak of one cycle to positive peak of the next cycle
3. From negative peak of one cycle to negative peak of the next cycle
The frequency of a wave is defined as the number
of cycles that a wave completes in one second.
In Fig. 4.4, the sine wave completes three
cycles in one second. Frequency is measured in
hertz. One hertz is equivalent to one cycle per
second, 60 hertz is 60 cycles per second, and so
on. In Fig. 4.4, the frequency denoted by f is 3
Hz, that is three cycles per second. The relation
between time period and frequency is given by
f =
Fig. 4.3
Fig. 4.4
1
T
A sine wave with a longer period consists of fewer cycles than one with a shorter period.
EXAMPLE 4.1
What is the period of the sine wave shown
in Fig. 4.5?
Solution From Fig. 4.5, it can be seen the
sine wave takes two seconds to complete
one period in each cycle
T 5 2s
Fig. 4.5
167
Introduction to Alternating Currents and Voltages
EXAMPLE 4.2
The period of a sine wave is 20 milliseconds. What is the frequency?
Solution
f =
=
1
T
1
= 50 Hz
20 ms
EXAMPLE 4.3
The frequency of a sine wave is 30 Hz. What is its period?
Solution
T=
=
1
f
1
= 0.03333 s = 33.33 ms
30
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 1*
Calculate the frequency of the following values of period.
(a) 0.2 s
(b) 50 ms
(c) 500 ms
(d) 10 ms
rrr 4-1.2 Calculate the period for each of the values of frequency.
(a) 60 Hz
(b) 500 Hz
(c) 1 kHz
(d) 200 kHz
rrr 4-1.1
4.2
(e) 5 MHz
ANGULAR RELATION OF A SINE WAVE
A sine wave can be measured along the X-axis on a time base which is
LO 2
frequency- dependent. A sine wave can also be expressed in terms of an
angular measurement. This angular measurement is expressed in degrees
or radians. A radian is defined as the angular distance measured along the
circumference of a circle which is equal to the radius of the circle. One
radian is equal to 57.3°. In a 360° revolution, there are 2p radians. The angular measurement of a sine wave
is based on 360° or 2p radians for a complete cycle as shown in Figs 4.6 (a) and (b).
V
V
Fig. 4.6
*Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
168
Circuits and Networks
A sine wave completes a half cycle in 180° or p radians; a quarter cycle in 90° or p/2 radians, and so on.
‰ Phase of a Sine Wave The phase of a sine wave is an angular measurement that specifies the position of
the sine wave relative to a reference. The wave shown in Fig. 4.7 is taken as the reference wave.
When the sine wave is shifted left or right with reference to the wave shown in Fig. 4.7, there occurs a
phase shift. Figure 4.8 shows the phase shifts of a sine wave.
In Fig. 4.8 (a), the sine wave is shifted to the right by 90° (p/2 rad) shown by the dotted lines. There is a phase
angle of 90° between A and B. Here, the waveform B is lagging behind waveform A by 90°. In other words, the
sine wave A is leading the waveform B by 90°. In Fig. 4.8 (b) the sine wave A is lagging behind the waveform B
by 90°. In both cases, the phase difference is 90°.
A
B
V
(volts)
90°
θ (degrees)
(a)
B A
V
(volts)
– 90°
θ (degrees)
(b)
Fig. 4.7
Fig. 4.8
EXAMPLE 4.4
What are the phase angles between the two sine waves shown in Figs 4.9 (a) and (b)?
A
V
(volts)
45° 90°
A B
V
(volts)
B
θ (degrees)
– 90°
30°
θ (degrees)
(b)
(a)
Fig. 4.9
Solution In Fig. 4.9 (a), the sine wave A is in phase with the reference wave; the sine wave B is out of phase,
which lags behind the reference wave by 45°. So we say that the sine wave B lags behind the sine wave A by
45°.
In Fig. 4.9 (b), the sine wave A leads the reference wave by 90°; the sine wave B lags behind the reference
wave by 30°. So the phase difference between A and B is 120°, which means that sine wave B lags behind sine
wave A by 120°. In other words, the sine wave A leads the sine wave B by 120°.
169
Introduction to Alternating Currents and Voltages
4.3
THE SINE WAVE EQUATION
LO 2
A sine wave is graphically represented as shown in Fig. 4.10 (a). The amplitude of a sine wave is represented
on vertical axis. The angular measurement (in degrees or radians) is represented on horizontal axis. Amplitude
A is the maximum value of the voltage or current on the Y-axis.
In general, the sine wave is represented by the equation
v(t) 5 Vmsin vt
The above equation states that any point on the sine wave represented by an instantaneous value v(t)
is equal to the maximum value times the sine of the angular frequency at that point. For example, if a
certain sine wave voltage has a peak value of 20 V, the instantaneous voltage at a point p/4 radians along the
horizontal axis can be calculated as
v(t ) = Vm sin vt
p
= 20 sin   = 20 × 0.707 = 14.14 V
 4 
When a sine wave is shifted to the left of the reference wave by a certain angle f, as shown in Fig. 4.10 (b),
the general expression can be written as
v(t) 5 Vmsin(vt 1 f)
When a sine wave is shifted to the right of the reference wave by a certain angle f, as shown in Fig.
4.10 (c), the general expression is
v(t) 5 Vmsin(vt 2 f)
Fig. 4.10
EXAMPLE 4.5
Determine the instantaneous value at the 90° point on the X-axis for each
sine wave shown in Fig. 4.11.
Fig. 4.11
170
Circuits and Networks
Solution From Fig. 4.11, the equation for the sine wave A
v(t) 5 10 sin vt
The value at p/2 in this wave is
p
v(t ) = 10 sin = 10 V
2
The equation for the sine wave B
v(t) 5 8 sin(vt 2 p/4)
vt 5 p/2
At
p p
v(t ) = 8 sin  − 
 2 4 
= 8 sin 45°
= 8 (0.707)
= 5.66 V
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 2
A certain sine wave has a positive going zero crossing at 0° and an rms value of 20 V. Calculate
its instantaneous value at each of the following angles.
(a) 33°
(b) 110°
(c) 145°
(d) 325°
rrr 4-2.2 For a particular 0° reference sinusoidal current, the peak value is 200 mA; determine the
instantaneous values at each of the following.
(a) 35°
(b) 190°
(c) 200°
(d) 360°
rrr 4-2.1
4.4
VOLTAGE AND CURRENT VALUES OF A SINE WAVE
As the magnitude of the waveform is not constant, the waveform can be
measured in different ways. These are instantaneous, peak, peak to peak, root
mean square (rms) and average
V (t) volts
10
values.
LO 3
4.2
2.1
6
6
1 2.5
7.5 10
–2
t (ms)
–8
– 10
Fig. 4.12
V(t)
Vp + 4V
t (ms)
– 4V
Fig. 4.13
‰Instantaneous Value Consider the sine wave shown in
Fig. 4.12. At any given time, it has some instantaneous value. This
value is different at different points along the waveform.
In Fig. 4.12, during the positive cycle, the instantaneous values
are positive and during the negative cycle, the instantaneous values
are negative. In Fig. 4.12 shown at time 1 ms, the value is 4.2 V; the
value is 10 V at 2.5 ms, – 2 V at 6 ms and – 10 V at 7.5, and so on.
‰Peak Value The peak value of the sine wave is the maximum
value of the wave during positive half cycle, or maximum value
of wave during negative half cycle. Since the values of these
two are equal in magnitude, a sine wave is characterised by a
single peak value. The peak value of the sine wave is shown in
Fig. 4.13; here the peak value of the sine wave is 4 V.
171
Introduction to Alternating Currents and Voltages
Peak-to-Peak Value The peak to peak value of a sine wave is the
value from the positive to the negative peak as shown in Fig. 4.14. Here,
the peak-to-peak value is 8 V.
V(t)
+ 4V
Average Value In general, the average value of any function v(t),
with period T is given by
1
vav =
T
t (ms)
– 4V
T
∫ v(t )dt
Fig. 4.14
0
That means that the average value of a curve in the X-Y plane is the total area under the complete curve
divided by the distance of the curve. The average value of a sine wave over one complete cycle is always zero.
So the average value of a sine wave is defined over a half-cycle, and not a full cycle period.
The average value of the sine wave is the total area under the half-cycle curve divided by the distance of
the curve.
The average value of the sine wave
v(t) 5 VP sin vt is given by
p
1
VP sin vt d (vt )
p ∫0
1
p
= [−VP cos vt ]0
p
2V
= P = 0.637 VP
Fig. 4.15
p
The average value of a sine wave is shown by the dotted line in Fig. 4.15.
vav =
EXAMPLE 4.6
Find the average value of a cosine wave f(t) = cos vt shown in Fig. 4.16.
Solution The average value of a cosine wave
v(t ) = VP cos vt
3p / 2
1
VP cos vt d (vt )
p p∫/ 2
1
= VP (− sin vt )3pp/ 2/ 2
p
−V
2V
= P [−1− 1] = P = 0.637 VP
p
p
Vav =
Fig. 4.16
Root Mean Square Value or Effective Value
The root mean square (rms) value of a sine wave is a
measure of the heating effect of the wave. When a resistor
is connected across a dc voltage source as shown in Fig.
4.17 (a), a certain amount of heat is produced in the
resistor in a given time. A similar resistor is connected
across an ac voltage source for the same time as shown
Fig. 4.17
172
Circuits and Networks
in Fig. 4.17 (b). The value of the ac voltage is adjusted such that the same amount of heat is produced in the
resistor as in the case of the dc source. This value is called the rms value.
That means the rms value of a sine wave is equal to the dc voltage that produces the same heating effect.
In general, the rms value of any function with period T has an effective value given by
Vrms =
1
T
T
2
∫ v(t ) dt
0
Consider a function v(t) 5 VP sin vt
The rms value, Vrms =
=
1
T
1
T
T
∫ (VP sin vt )
2
d ( vt )
0
2p
1− cos 2vt 
 d ( vt )
2

∫ VP 
2
0
VP
= 0.707 VP
2
If the function consists of a number of sinusoidal terms, that is
=
v(t ) = V0 + (Vc1 cos vt + Vc 2 cos 2 vt + �) + (Vs1 sin vt + Vs 2 sin 2 vt + �)
The rms, or effective value, is given by
1
1
Vrms = V02 + (Vc21 + Vc22 + �) + (Vs21 + Vs22 + �)
2
2
EXAMPLE 4.7
A wire is carrying a direct current of 20 A and a sinusoidal alternating current of peak value 20 A. Find the
rms value of the resultant current in the wire.
Solution The rms value of the combined wave
= 20 2 +
20 2
2
= 400 + 200 = 600 = 24.5 A
Peak Factor The peak factor of any waveform is defined as the ratio of the peak value of the wave to
the rms value of the wave.
Peak factor =
VP
Vrms
Peak factor of the sinusoidal waveform =
VP
VP / 2
= 2 = 1.414
173
Introduction to Alternating Currents and Voltages
‰ Form Factor Form factor of a waveform is defined as the ratio of rms value to the average value of
the wave.
Form factor =
Vrms
Vav
Form factor of a sinusoidal waveform can be found from the above relation.
For the sinusoidal wave, the form factor =
VP / 2
= 1.11
0.637VP
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 3
rrr 4-3.1
rrr 4-3.2
rrr 4-3.3
rrr 4-3.4
rrr 4-3.5
rrr 4-3.6
Sine wave A lags the sine wave B by 30°. Both have peak values of 15 V. Sine wave A is the
reference with a positive going crossing at 0°. Determine the instantaneous value of the sine wave
B at 30°, 90°, 45°, 180° and 300°.
Find the rms values of
(a) v (t) 5 25 cos vt 1 15 sin vt
(b) i(t) 5 100 sin vt 2 10 cos 2vt
A sawtooth voltage wave increases linearly
from 0 to 200 V in the interval from 0 to 2
seconds. At t1 5 2 s, its value drops to zero
suddenly. The waveform repeats this pattern.
Find the rms value of the voltage wave.
Determine the Vrms of the waveform shown in
Fig. Q.4
Fig. Q.4.
Find the effective value of the resultant current in
a wire which carries a direct current of 10 A and a
sinusoidal current with a peak value of 15 A.
Determine the value of K in the waveform
shown in Fig. Q.6 where K is some function
Fig. Q.6
of the period T such that the effective value
is 2.
rrr 4-3.7
Determine the average and rms values of
the waveform shown in Fig. Q.7, where in
the first interval of v(t) 5 20 e–200t.
rrr 4-3.8
Find the effective value of the function
v 5 100 1 50 sin vt.
rrr 4-3.9
A full-wave rectified sine wave is clipped at
0.707 of its maximum value as shown in Fig.
Q.9. Find the average and effective values of
the function.
Fig. Q.7
Fig. Q.9
174
Circuits and Networks
rrr4-3.10 Find
the rms value of the function shown in
Fig. Q.10 and described as follows:
0 , t , 0.1v 5 40 (1 2 e–100t)
0.1 , t , 0.2v 5 40 e –50(t 2 0.1)
rrr 4-3.11 Calculate average and effective values of the
Fig. Q.10
waveform shown in Fig. Q.11 and, hence, find
from factor.
rrr 4-3.12 A full-wave rectified sine wave is clipped such that the effective value is 0.5 Vm as shown in Fig.
Q.12. Determine the amplitude at which the wave form is clipped.
Fig. Q.11
Fig. Q.12
Frequently Asked Questions linked to LO 3
rrr 4-3.1
Define RMS voltage.
rrr4-3.2 A periodic voltage waveform has been shown in Fig. Q.14.
Determine the following. [JNTU Nov. 2012]
(a) Frequency of the waveform
(b) Wave equation for 0 < t < 100 ms
(c) R.M.S value and
(d) Average value
rrr 4-3.3
[AU May/June 2014]
V (t)
Vm
0
p
2p
t (ms)
Fig. Q.14
A non-alternating periodic waveform has been show in Fig. Q.3.
Find its form factor and peak factor.
[JNTU Nov. 2012]
x(t)
2A
1A
0
10
20
30
40
Fig. Q.3
rrr 4-3.4
Explain the rms value and average value of alternating quantity. Derive its necessary expressions.
[JNTU Nov. 2012]
rrr 4-3.5 Define the following
[JNTU Nov. 2012]
(a) Time period
(b) Frequency
(c) RMS value
(d) Average value
4.5
PHASE RELATION IN A PURE RESISTOR
When a sinusoidal voltage of certain magnitude is applied to a resistor, a certain
amount of sine wave current passes through it. We know the relation between
v (t) and i(t) in the case of a resistor. The voltage/current relation in case of a
resistor is linear,
i.e.
v (t) 5 i(t)R
LO 4
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
Introduction to Alternating Currents and Voltages
175
Consider the function
i(t ) = I m sin vt = IM  I m e jvt  or I m ∠0°


If we substitute this in the above equation, we have
v(t ) = I m R sin vt =Vm sin vt
= IM Vm e jvt  or Vm∠0°


Vm = I m R
where
If we draw the waveform for both voltage and current as shown in Fig. 4.18, there is no phase difference
between these two waveforms. The amplitudes of the waveform may differ according to the value of resistance.
As a result, in pure resistive circuits, the voltages and currents are
said to be in phase. Here the term impedance is defined as the ratio of
voltage to current function. With ac voltage applied to elements, the ratio
of exponential voltage to the corresponding current (impedance) consists of
magnitude and phase angles. Since the phase difference is zero in case of a
resistor, the phase angle is zero. The impedance in case of resistor consists
only of magnitude, i.e.
Fig. 4.18
4.6
Z=
Vm∠0°
=R
I m∠0°
PHASE RELATION IN A PURE INDUCTOR
LO 4
As discussed earlier in Chapter 1, the voltage current relation in the case of an inductor is given by
di
v (t ) = L
dt
j vt
Consider the function i(t ) = I m sin vt = IM  I m e  or I m∠0°
d
v (t ) = L ( I m sin vt )
dt
= Lv I m cos vt = vL I m cos vt
v (t ) = Vm cos vt , or Vm sin (vt + 90°)
o
= IM Vm e j ( vt +90 )  or Vm∠90°


where Vm 5 vLIm 5 XLIm
and e j90° = j = 1∠90°
If we draw the waveforms for both, voltage and current, as shown in Fig.
4.19, we can observe the phase difference between these two waveforms.
As a result, in a pure inductor the voltage and current are out of
phase. The current lags behind the voltage by 90° in a pure inductor as
shown in Fig. 4.20.
Fig. 4.19
176
Circuits and Networks
The impedance which is the ratio of exponential voltage to the corresponding current, is given by
Vm sin (vt + 90°)
I m sin vt
where Vm = vL I m
I vL sin (vt + 90°)
= m
I m sin vt
vL I m∠90°
=
I m∠0°
∴ Z = jvL = jX L
Z=
Fig. 4.20
where XL 5 vL and is called the inductive reactance.
Hence, a pure inductor has an impedance whose value is vL.
4.7
PHASE RELATION IN A PURE CAPACITOR
LO 4
As discussed in Chapter 1, the relation between voltage and current is given by
1
v(t ) = ∫ i (t ) dt
C
Consider the function i(t ) = I m sin vt = IM  I m e jvt  or I m∠0°


1
v(t ) = ∫ I m sin vt d (t )
C
1
I m [− cos vt ]
=
vC
Im
sin (vt − 90o )
vC
∴ v(t ) = Vm sin (vt − 90o )
=
o
= IM  I m e j ( vt −90 )  or Vm∠− 90o


Im
where
Vm =
vC
Vm∠− 90o
−j
∴
=Z=
o
vC
I m∠0
−j
Hence, the impedance is Z =
= − jX C
vC
1
where X C =
and is called the capacitive reactance.
vC
If we draw the waveform for both, voltage and current, as shown
in Fig. 4.21, there is a phase difference between these two waveforms.
As a result, in a pure capacitor, the current leads the voltage by 90°.
The impedance value of a pure capacitor
1
XC =
Fig. 4.21
vC
177
Introduction to Alternating Currents and Voltages
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 4
r 4-4.1
A sinusoidal voltage of v(t) 5 50 sin (500t) applied to a capacitive circuit. Determine the capacitive
reactance, and the current in the circuit.
Additional Solved Problems
PROBLEM 4.1
Find the average values of the voltages across R1 and R2. In
Fig. 4.22, values shown are rms.
Solution The voltage across the 2 V resistor V2 5 30 V
The voltage across the 5 V resistor V5 5 50 V
Fig. 4.22
Peak value of the voltage across the 2 V resistor
2 ×30 = 42.43 V
Peak value of the voltage across the 5 V resistor
2 ×50 = 70.71 V
Average value across the 2 V resistor 5 42.43 3 0.637 5 27.03 V
Average value across the 5 V resistor 5 70.71 3 0.637 5 45 V
PROBLEM 4.2
A sinusoidal voltage is applied to the circuit shown in Fig. 4.23, determine
rms current, average current, peak current, and peak-to-peak current.
Solution The equation for the applied voltage v(t) 5 Vpsin vt
5 10 sin vt
Vp
sin vt = 10 ×10−3 sin vt
The equation for the current i(t ) =
1kV
Peak value of the current ip 5 10 mA
10
rms value of the current irms =
= 7.071 mA
2
Peak-to-peak value of the current ipp 5 20 mA
Average value of the current iav 5 10 3 0.637 5 6.37 mA.
1 kΩ
Vp
10 V
Fig. 4.23
PROBLEM 4.3
A sinusoidal voltage source in series with a dc source as shown in Fig. 4.24. Sketch the voltage across RL.
Determine the maximum current through RL and the average voltage across RL.
Fig. 4.24
178
Circuits and Networks
Solution The voltage equation v(t) 5 150 sin vt
The voltage across resistance RL is
vRL (t ) = 200 + 150 sin vt V.
Fig. 4.25
Peak-to-peak value of the voltage is 300 V.
The voltage across RL is shown in Fig. 4.25.
The current through the resistance RL is
1
iRL (t ) =
[200 + 150 sin vt ]
100
iRL (t ) = 2 + 1.5 sin vt A
Hence the maximum current is 3.5 A
Average voltage across RL 5 200 V.
PROBLEM 4.4
An alternating current varying sinusoidally, with a frequency of 50 Hz has an rms value of 20 A. Write down the
equation for the instantaneous value and find this value at (a) 0.0025 s, and (b) 0.0125 s after passing through
a positive maximum value. At what time, measured from a positive maximum value, will the instantaneous
current be 14.14 A?
Solution The rms value of the current waveform
Irms 5 20A
Peak of the current waveform
I m = 2 × 20 = 28.28 volts
The current equation i(t) 5 Imsin vt 5 28.28 sin vt
If it is passing through maximum positive value
i(t) 5 28.28 cos vt 5 28.28 cos 2pft
i(t) 5 28.28 cos100 pt
(a)
(b)
At t =
At t 5 0.0025 s
i(t) 5 28.28 cos100 p (0.0025) 5 20 A
At t 5 0.0125 s
i(t) 5 28.28 cos100 p (0.0125) 5 – 20 A
1
s, the instantaneous current becomes 14.14 A.
300
PROBLEM 4.5
Determine the rms value of the voltage defined by

p
v(t ) = 5 + 5 sin 314t + 

6
Solution If the function consists of a number of sinusoidal terms, i.e.
v (t) 5 V0 1 (VC1 cos vt 1 VC2 cos 2 vt 1 ...) 1 (VS1 sin vt 1 VS2 sin 2 vt 1 ...)
The rms, or effective value is given by
1
Vrms = V0 2 + (VC12 + VC 2 2 + ...) + (VS12 + VS 2 2 + ...)
2
179
Introduction to Alternating Currents and Voltages
From the equation

p
v(t ) = 5 + 5 sin 314 t + 

6
The rms value of the waveform is
Vrms = 52 +
52
= 25 + 12.5 = 6.12 V.
2
PROBLEM 4.6
Sine wave ‘A’ has a positive going zero crossing at 45°. Sine wave ‘B’ has
a positive going zero crossing at 60°. Determine the phase angle between
the signals. Which of the signals lags behind the other?
Solution The two signals are shown in Fig. 4.26.
From Fig. 4.26, the signal B lags behind signal A by 15°. In other
words, the signal A leads the signal B by 15°.
Fig. 4.26
PROBLEM 4.7
One sine wave has a positive peak at 75°, and another has a positive peak at 100°. How much is each sine
wave shifted in phase from the 0° reference? What is the phase angle between them?
Solution The two signals are drawn as shown in Fig. 4.27.
Fig. 4.27
The signal A leads the reference signal by 15°.
The signal B lags behind the reference signal by 10°.
The phase angle between these two signals is 25°.
PROBLEM 4.8
A sinusoidal voltage is applied to the resistive circuit shown in Fig. 4.28.
Determine the following values.
(a)
(c)
Irms
IP
(b)
(d)
Iav
IPP
Solution The function given to the circuit shown is
v(t) 5 VP sin vt 5 20 sin vt
Fig. 4.28
180
Circuits and Networks
The current passing through the resistor
i (t ) =
i (t ) =
v (t )
R
20
2×103
sin vt
= 10 ×10−3 sin vt
I P = 10 ×10−3 A
IP 5 10 mA
The peak value
Peak-to-peak value IPP 5 20 mA
rms value Irms
5 0.707 IP
5 0.707 3 10 mA 5 7.07 mA
Average value Iav
5 (0.637) IP
5 0.637 3 10 mA 5 6.37 mA
PROBLEM 4.9
A sinusoidal voltage is applied to a capacitor as shown in Fig. 4.29. The frequency
of the sine wave is 2 kHz. Determine the capacitive reactance.
Solution
XC =
=
1
2pfC
1
2p× 2×103 × 0.01×10−6
Fig. 4.29
= 7.96 kV
PROBLEM 4.10
Determine the rms current in the circuit shown in Fig. 4.30.
1
XC =
Solution
2p fC
1
=
3
2p×5×10 × 0.01×10−6
= 3.18 kV
V
5
I rms = rms =
=1.57 mA
X C 3.18 K
Fig. 4.30
PROBLEM 4.11
A sinusoidal voltage is applied to the circuit shown in Fig. 4.31. The frequency
is 3 kHz. Determine the inductive reactance.
Solution XL 5 2pfL
5 2p 3 3 3 103 3 2 3 10–3
5 37.69 V
Fig. 4.31
181
Introduction to Alternating Currents and Voltages
PROBLEM 4.12
Determine the rms current in the circuit shown in Fig. 4.32.
Solution XL 5 2pfL
5 2p 3 10 3 103 3 50 3 10–3
XL 5 3.141 kV
V
I rms = rms
XL
10
=
= 3.18 mA
3.141×103
Fig. 4.32
PROBLEM 4.13
Find the form factor of the half-wave rectified sine
wave shown in Fig. 4.33.
Solution v 5 Vm sin vt,
for 0 , vt , p
for p , vt , 2p
5 0,
the period is 2p.
Average value
Fig. 4.33
2p

p
1 


V
sin
t
d
(
t
)
+
d
(
t
)
v
v
0
v
∫ m

∫

2p 

p


0
= 0.318Vm
Vav =
p
1
1
2
(Vm sin vt ) d (vt ) = Vm2
∫
2p 0
4
1
Vrms = Vm
2
Vrms
=
Vav
2
=
Vrms
Form factor
=
0.5Vm
0.318Vm
= 1.572
PROBLEM 4.14
Find the average and effective values of the sawtooth wave-form
shown in Fig. 4.34 below.
Solution From Fig. 4.34 shown, the period is T.
1
Vav =
T
=
T
∫
0
Vm
1 Vm
t dt =
T
T T
2
V
Vm T
= m
2 2
2
T
T
∫
0
t dt
Fig. 4.34
182
Circuits and Networks
Effective value
1
=
T
=
T
1
T
Vrms =
∫
v 2 dt
0
T
2
Vm

 T
∫
0

t  dt

Vm
3
PROBLEM 4.15
Find the average and rms value of the full-wave rectified sine wave
shown in Fig. 4.35.
p
Solution
1
5sin vt d (vt )
p ∫0
= 3.185
Average value Vav =
Fig. 4.35
p
1
2
(5sin vt ) d (vt )
p ∫0
Effective value or rms value =
25
= 3.54
2
=
PROBLEM 4.16
The full-wave rectified sine wave shown in Fig. 4.36 has a delay
angle of 60°. Calculate Vav and Vrms.
p
Solution Average value Vav =
1
10 sin (vt ) d (vt )
p ∫0
1
=
p
Vav =
Effective value Vrms=
=
1
p
p
∫
10 sin vt d (vt )
60o
10
p
(− cos vt )60
= 4.78
p
p
∫ (10 sin vt )
2
d ( vt )
60o
100
p
= 6.33
p
1− cos 2vt 
 d (vt )

2
o
∫ 
60
Fig. 4.36
183
Introduction to Alternating Currents and Voltages
PROBLEM 4.17
Find the form factor of the square wave as shown in Fig. 4.37.
Solution v 5 20 for
0 , t , 0.01
5 0 for 0.01 , t , 0.03
The period is 0.03 second.
Average value Vav =
=
Effective value Veff =
1
0.03
Fig. 4.37
0.01
∫
20 dt
0
20 (0.01)
= 6.66
0.03
1
0.03
0.01
∫
( 20) 2 dt = 66.6 = 0.816
0
0.816
Form factor =
= 0.123
6.66
Answers to Practice Problems
4-1.1 5 Hz: 20 Hz; 2 kHz; 100 kHz
4-2.1 15.4 V; 26.57 V; 16.22 V; 216.22 V
4-3.1 12.99 V; 12.99 V; 14.49 V; 27.5 V; 27.5 V
4-3.2 Vrms 5 20.62 V; Irms 5 71.06 V
4-3.3 Vrms 5 115.47 volts
4-3.4
Vrms 5
4-3.6
K 5 0.12
4-3.7
Vav 5 2 V; Vrms 5 4.47 V
4-3.10
27.57
4-3.12
0.581 Vm or 35.5°
Vm
3
Objective-Type Questions
rrr 4.1
rrr 4.2
rrr 4.3
rrr 4.4
rrr 4.5
One sine wave has a period of 2 ms, another has a period of 5 ms, and yet other has a period of 10 ms. Which
sine wave is changing at a faster rate?
(a) Sine wave with period 2 ms
(b) Sine wave with period of 5 ms
(c) All are at the same rate
(d) Sine wave with period of 10 ms
How many cycles does a sine wave go through in 10 s when its frequency is 60 Hz?
(a) 10 cycles
(b) 60 cycles
(c) 600 cycles
(d) 6 cycles
If the peak value of a certain sine wave voltage is 10 V, what is the peak-to- peak value?
(a) 20 V
(b) 10 V
(c) 5 V
(d) 7.07 V
If the peak value of a certain sine wave voltage is 5 V, what is the rms value?
(a) 0.707 V
(b) 3.535 V
(c) 5 V
(d) 1.17 V
What is the average value of a sine wave over a full cycle?
(a) Vm
(b)
Vm
2
(c) zero
(d)
2Vm
184
rrr 4.6
rrr 4.7
rrr 4.8
rrr 4.9
rrr 4.10
rrr 4.11
rrr 4.12
rrr 4.13
rrr 4.14
rrr 4.15
rrr 4.16
rrr 4.17
rrr 4.18
rrr 4.19
rrr 4.20
rrr 4.21
rrr 4.22
Circuits and Networks
A sinusoidal current has peak value of 12 A. What is its average value?
(a) 7.64 A
(b) 24 A
(c) 8.48 A
(d) 12 A
Sine wave A has a positive going zero crossing at 30°. Sine wave B has a positive going zero crossing at 45°.
What is the phase angle between two signals?
(a) 30°
(b) 45°
(c) 75°
(d) 5°
A sine wave has a positive going zero crossing at 0° and an rms value of 20 V. What is its instantaneous value
at 145°.
(a) 7.32 V
(b) 16.22 V
(c) 26.57 V
(d) 21.66 V
In a pure resistor, the voltage and current are
(a) out of phase
(b) in phase
(c) 90° out of phase
(d) 45° out of phase
The rms current through a 10 kV resistor is 5 mA. What is the rms voltage drop across the resistor?
(a) 10 V
(b) 5 V
(c) 50 V
(d) zero
In a pure capacitor, the voltage
(a) is in phase with the current
(b) is out of phase with the current
(c) lags behind the current by 90°
(d) leads the current by 90°
A sine-wave voltage is applied across a capacitor; when the frequency of the voltage is increased, the current
(a) increases
(b) decreases
(c) remains the same
(d) is zero
The current in a pure inductor
(a) lags behind the voltage by 90°
(b) leads the voltage by 90°
(c) is in phase with the voltage
(d) lags behind the voltage by 45°
A sine-wave voltage is applied across an inductor; when the frequency of voltage is increased, the current
(a) increases
(b) decreases
(c) remains the same
(d) is zero
The rms value of the voltage for a voltage function v 5 10 1 5 cos (628t 1 30°) volts through a circuit is
(a) 5 V
(b) 10 V
(c) 10.6 V
(d) 15 V
For the same peak value, which of the following waves will have the highest rms value?
(a) Sine wave
(b) Square wave
(c) Triangular wave
(d) Half-wave rectified sine wave
For 100 volts rms value triangular wave, the peak voltage will be
(a) 100 V
(b) 111 V
(c) 141 V
(d) 173 V
v
The form factor of dc voltage is
(a) zero
(b) infinite (c) unity (d) 0.5
Vm
For the half-wave rectified sine wave shown in Fig. 4.38,
the peak factor is
(a) 1.41
(b) 2.0
(c) 2.82
(d) infinite
t
2
3
4
0
For the square wave shown in Fig. 4.39, the form
Fig. 4.38
factor is
(a) 2.0
(b) 1.0
(c) 0.5
(d) zero
The power consumed in a circuit element will be least when the
phase difference between the current and voltage is
(a) 0°
(b) 30°
(c) 90°
(d) 180°
A voltage wave consists of two components: a 50 V dc
component and a sinusoidal component with a maximum value
Fig. 4.39
of 50 volts. The average value of the resultant will be
(a) zero
(b) 86.6 V
(c) 50
(d) none of the above
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5
LEARNING OBJECTIVES
5.1
IMPEDANCE DIAGRAM
So far, our discussion has been confined to resistive circuits. Resistance
LO 1
restricts the flow of current by opposing free electron movement. Each
element has some resistance; for example, an inductor has some resistance; a
capacitance also has some resistance. In the resistive element, there is no phase
difference between the voltage and the current. In the case of pure inductance, the current lags behind
the voltage by 90 degrees, whereas in the case of a pure capacitance, the current leads the voltage by 90
degrees. Almost all electric circuits offer impedance to the flow of current. Impedance is a complex quantity
having real and imaginary parts; where the real part is the resistance and the imaginary part is the reactance
of the circuit.
Consider the RL series circuit shown in Fig. 5.1. If we apply the real function Vm cos vt to the circuit,
the response may be Im cos vt. Similarly, if we apply the imaginary function jVm sin vt to the same circuit,
the response is jIm sin vt. If we apply a complex function, which is a combination of real and imaginary
functions, we will get a complex response.
This complex function is Vm e jvt 5 Vm (cos vt 1 j sin vt).
Applying Kirchhoff’s law to the circuit shown in Fig. 5.1,
di(t )
we get
Vm e jvt = Ri(t ) + L
dt
The solution of this differential equation is
i(t) 5 Im ejvt
Vme jwt
Fig. 5.1
By substituting i(t) in the above equation, we get
d
Vm e jvt = R I m e jvt + L ( I m e jvt )
dt
186
Circuits and Networks
Vm e jvt 5 RIm e jvt 1 LIm jve jvt
Vm 5 (R 1 jvL)Im
Impedance is defined as the ratio of the voltage to current function
Z=
Vm e jvt
= R + j vL
Vm
e j vt
R + j vL
Complex impedance is the total opposition offered by the circuit elements
to ac current, and can be displayed on the complex plane. The impedance is
denoted by Z. Here the resistance R is the real part of the impedance, and the
reactance XL is the imaginary part of the impedance. The resistance R is located
on the real axis. The inductive reactance XL is located on the positive j axis. The
resultant of R and XL is called the complex impedance.
Figure 5.2 is called the impedance diagram for the RL circuit. From Fig.
5.2, the impedance Z = R2 + (vL)2 , and angle u 5 tan21 vL/R. Here, the
impedance is the vector sum of the resistance and inductive reactance. The
angle between impedance and resistance is the phase angle between the current
and voltage applied to the circuit.
Similarly, if we consider the RC series circuit, and apply the complex
function Vm e jvt to the circuit in Fig. 5.3, we get a complex response as follows.
Applying Kirchhoff’s law to the above circuit, we get
Fig. 5.2
Vm e jvt = Ri(t ) +
Fig. 5.3
1
C
∫
i(t ) dt
Solving this equation, we get
i(t) 5 Im e jvt
Vm e jvt = R I m e jvt +
 +1
1
I m   e jvt
C
 jv 


j
=  RI m −
I m  e j vt

vC 

j 
Vm =  R −
 Im

vC 
The impedance
Z=
Vm e jvt
Vm / [ R − j / vC ] e jvt
5 [R 2 ( j/vC)]
Fig. 5.4
Here, the impedance Z consists of resistance (R), which is the real part, and
capacitive reactance (XC 5 1/ vC), which is the imaginary part of the impedance.
Complex Impedance
187
The resistance, R, is located on the real axis, and the capacitive reactance XC is located on the negative j axis
in the impedance diagram in Fig. 5.4.
From Fig. 5.4, impedance Z = R2 + X C2 or R2 + (1 / vC )2 and angle u 5 tan21 (1/vCR). Here, the
impedance, Z, is the vector sum of resistance and capacitive reactance. The angle between resistance and
impedance is the phase angle between the applied voltage and current in the circuit.
5.2
PHASOR DIAGRAM
A phasor diagram can be used to represent a sine wave in terms of its magnitude
LO 2
and angular position. Examples of phasor diagrams are shown in Fig. 5.5.
In Fig. 5.5 (a), the length of the arrow represents the magnitude of the sine
wave; angle u represents the angular position of the sine wave. In Fig. 5.5 (b),
the magnitude of the sine wave is one and the phase angle is 30°. In Fig. 5.5 (c)
and (d), the magnitudes are four and three, and phase angles are 135° and 225°, respectively. The position
of a phasor at any instant can be expressed
as a positive or negative angle. Positive
angles are measured counterclockwise from
0°, whereas negative angles are measured
clockwise from 0°. For a given positive
angle u, the corresponding negative angle
is u 2 360°. This is shown in Fig. 5.6 (a).
In Fig. 5.6 (b), the positive angle 135° of
vector A can be represented by a negative
angle 2225°, (135° 2 360°).
A phasor diagram can be used to represent
the relation between two or more sine
waves of the same frequency. For example,
the sine waves shown in Fig. 5.7 (a) can be
represented by the phasor diagram shown in
Fig. 5.7 (b).
In the above figure, the sine wave B lags
Fig. 5.5
behind the sine wave A by 45°; the sine wave
C leads the sine wave A by 30°. The length of
the phasors can be used to represent peak, rms, or average values.
Fig. 5.6
188
Circuits and Networks
Fig. 5.7
EXAMPLE 5.1
Draw the phasor diagram to represent the two sine waves
shown in Fig. 5.8.
Solution The phasor diagram representing the sine waves is
shown in Fig. 5.9. The length of the each phasor represents the
peak value of the sine wave.
Fig. 5.8
Fig. 5.9
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 2
rrr 5-2.1
A 250 V, 50 Hz voltage is applied to a coil of 5 H inductance and resistance of 2 V in series with
a capacitance C. What value must ‘C’ have in order that the voltage across the coil shall be 280 V?
Draw the phasor diagram.
*Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
189
Complex Impedance
5.3
SERIES CIRCUITS
The impedance diagram is a useful tool for analysing series ac circuits.
Basically we can divide the series circuits as RL, RC, and RLC circuits.
In the analysis of series ac circuits, one must draw the impedance
diagram. Although the impedance diagram usually is not drawn to
scale, it does represent a clear picture of the phase relationships.
LO 3
5.3.1 Series RL Circuit
If we apply a sinusoidal input to an RL circuit, the current in the circuit and all voltages across the elements
are sinusoidal. In the analysis of the RL series circuit, we can find the impedance, current, phase angle and
voltage drops. In Fig. 5.10 (a), the resistor voltage (VR) and
current (I ) are in phase with each other, but lag behind the
source voltage (VS). The inductor voltage (VL) leads the
source voltage (VS). The phase angle between current and
voltage in a pure inductor is always 90°. The amplitudes
of voltages and currents in the circuit are completely
dependent on the values of elements (i.e. the resistance and
inductive reactance). In the circuit shown, the phase angle
Fig. 5.10 (a)
is somewhere between zero and 90° because of the series
combination of resistance with inductive reactance, which
V
VL
depends on the relative values of R and XL.
VR
The phase relation between current and voltages in a
series RL circuit is shown in Fig. 5.10 (b).
I
Here, VR and I are in phase. The amplitudes are arbitrarily
chosen.
From Kirchhoff’s voltage law, the sum of the
0
θ
voltage drops must equal the applied voltage. Therefore,
the source voltage VS is the phasor sum of VR and VL.
90º
Fig. 5.10 (b)
∴
Vs = V 2R + V 2L
The phase angle between resistor voltage and source voltage is
u 5 tan21 (VL/VR)
where u is also the phase angle between the source voltage and the current. The
phasor diagram for the series RL circuit that represents the waveforms in Fig.
5.10 (c).
EXAMPLE 5.2
Fig. 5.10 (c)
To the circuit shown in Fig. 5.11, consisting a 1 kV resistor
connected in series with a 50 mH coil, a 10 V rms, 10 kHz signal
is applied. Find impedance Z, current I, phase angle u, voltage
across resistance VR , and the voltage across inductance VL.
Fig. 5.11
190
Circuits and Networks
Solution Inductive reactance XL 5 vL
5 2pf L 5 (6.28) (104) (50 3 1023) 5 3140 V
In rectangular form,
Total impedance Z 5 (1000 1 j3140) V
R2 + X L2
(1000)2 + (3140)2 = 3295.4 V
Current I 5 VS /Z 5 10/3295.4 5 3.03 mA
Phase angle u 5 tan21 (XL/R) 5 tan21 (3140/1000) 5 72.33°
Therefore, in polar form, total impedance Z 5 3295.4 ∠72.33°
Voltage across resistance VR 5 IR
5 3.03 3 1023 3 1000 5 3.03 V
Voltage across inductive reactance VL 5 IXL
5 3.03 3 1023 3 3140 5 9.51 V
EXAMPLE 5.3
Determine the source voltage and the phase angle, if voltage across the
resistance is 70 V and voltage across the inductive reactance is 20 V as
shown in Fig. 5.12.
Solution In Fig. 5.12, the source voltage is given by
VS = VR2 +VL2
Fig. 5.12
= (70) + (20) = 72.8 V
2
2
The angle between current and source voltage is
u 5 tan21 (VL/VR) 5 tan21 (20/70) 5 15.94°
5.3.2 Series RC Circuit
When a sinusoidal voltage is applied to an RC series circuit, the current in the circuit and voltages across each
of the elements are sinusoidal. The series RC circuit is shown in Fig. 5.13 (a).
Here, the resistor voltage and current are in phase with each
other. The capacitor voltage lags behind the source voltage. The
phase angle between the current and the capacitor voltage is
always 90°. The amplitudes and the phase relations between the
voltages and current depend on the ohmic values of the resistance
and the capacitive reactance. The circuit is a series combination
of both resistance and capacitance; and the phase angle between
Fig. 5.13 (a)
the applied voltage and the total current is somewhere between
zero and 90°, depending on the relative values of the resistance
191
Complex Impedance
and reactance. In a series RC circuit, the current is the same through the resistor and the capacitor. Thus, the
resistor voltage is in phase with the current, and the capacitor voltage lags behind the current by 90° as shown
in Fig. 5.13 (b).
Here, I leads VC by 90°. VR and I are in phase. From Kirchhoff’s voltage law, the sum of the voltage drops
must be equal to the applied voltage. Therefore, the source voltage is given by
VS = VR2 +VC2
The phase angle between the resistor voltage and the
source voltage is
u 5 tan21 (VC /VR)
Fig. 5.13 (b)
Since the resistor voltage and the current are in phase, u
also represents the phase angle between the source voltage
and current. The voltage phasor diagram for the series RC
circuit, voltage, and current phasor diagrams represented by
the waveforms in Fig. 5.13 (b) are shown in Fig. 5.13 (c).
Fig. 5.13 (c)
EXAMPLE 5.4
A sine-wave generator supplies a 500 Hz, 10 V rms signal to
a 2 kV resistor in series with a 0.1 mF capacitor as shown in
Fig. 5.14. Determine the total impedance Z, current I, phase
angle u, capacitive voltage VC , and resistive voltage VR .
Solution To find the impedance Z, we first solve for XC
XC =
1
1
=
2pfC 6.28×500×0.1×10−6
5 3184.7 V
In rectangular form,
Total impedance Z 5 (2000 2 j3184.7) V
Z = (2000)2 + (3184.7)2
5 3760.6 V
Phase angle u 5 tan21 (2XC /R) 5 tan21(23184.7/2000) 5 257.87°
Fig. 5.14
192
Circuits and Networks
Current I 5 VS /Z 5 10/3760.6 5 2.66 mA
Capacitive voltage VC 5 IXC
5 2.66 3 1023 3 3184.7 5 8.47 V
Resistive voltage VR 5 IR
5 2.66 3 1023 3 2000 5 5.32 V
The arithmetic sum of VC and VR does not give the applied voltage of 10 volts. In fact, the total applied
voltage is a complex quantity. In rectangular form,
Total applied voltage VS 5 5.32 2 j8.47 V
In polar form,
VS 5 10 ∠257.87° V
The applied voltage is complex, since it has a phase angle relative to the resistive current.
EXAMPLE 5.5
Determine the source voltage and phase angle when the voltage across
the resistor is 20 V and the capacitor is 30 V as shown in Fig. 5.15.
Solution Since VR and VC are 90° out of phase, they cannot be added
directly. The source voltage is the phasor sum of VR and VC .
∴ VS = VR2 + VC2 = (20)2 + (30)2 = 36 V
Fig. 5.15
The angle between the current and source voltage is
u 5 tan–1 (VC /VR) 5 tan–1 (30/20) 5 56.3°
5.3.3 Series RLC Circuit
A series RLC circuit is the series combination
of resistance, inductance and capacitance. If
we observe the impedance diagrams of series
RL and series RC circuits as shown in Fig.
5.16 (a) and (b), the inductive reactance, XL ,
is displayed on the 1 j axis and the capacitive
reactance, XC , is displayed on the – j axis.
These reactance are 180° apart and tend to
cancel each other.
Fig. 5.16
The magnitude and type of reactance in a
series RLC circuit is the difference of the two reactance. The impedance for an RLC series circuit is given by
Z = R2 + ( X L − X C )2 . Similarly, the phase angle for an RLC circuit is
 X − X C 
= tan−1  L


R

193
Complex Impedance
EXAMPLE 5.6
In the circuit shown in Fig. 5.17, determine the total
impedance, current I, phase angle u, and the voltage
across each element.
Solution To find impedance Z, we first solve for XC and
XL
XC =
1
1
=
2pfC 6.28×50×10×10−6
Fig. 5.17
5 318.5 V
XL 5 2pfL 5 6.28 3 0.5 3 50 5 157 V
Total impedance in rectangular form
Z 5 (10 1 j157 – j 318.5) V
5 10 1 j(157 – 318.5) V 5 10 – j161.5 V
Here, the capacitive reactance dominates the inductive reactance.
Z = (10) 2 + (161.5) 2
= 100 + 26082.2 =161.8 V
50
= 0.3 A
161.8
Phase angle u 5 tan–1 [(XL 2 XC)/R] 5 tan–1 (2161.5/10) 5 2 86.45°
Current
I =VS Z =
Voltage across the resistor VR 5 IR 5 0.3 3 10 5 3 V
Voltage across the capacitive reactance 5 IXC 5 0.3 3 318.5 5 95.55 V
Voltage across the inductive reactance 5 IXL 5 0.3 3 157 5 47.1 V
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 3
r 5-3.1
For the circuit shown in Fig. Q.1, determine the impedance, phase angle, and total current.
Fig. Q.1
194
Circuits and Networks
rrr 5-3.2
Determine the impedance and phase angle in the circuit shown in Fig. Q.2.
rrr 5-3.3
Calculate the impedance at each of the following frequencies; also determine the current at each
frequency in the circuit shown in Fig. Q.3.
(a) 100 Hz
(b) 3 kHz
Fig. Q.2
Fig. Q.3
rrr 5-3.4
Find the values of R and C in the circuit shown in Fig. Q.4 so that
Vb 5 4Va and Va, Vb are in phase quardrature.
rrr 5-3.5
The applied voltage to a series circuit is v(t) 5 50 sin (2000t225°)
volts and the resultant current through the circuit is i(t) 5 8 sin
(2000t 1 5°) amperes. Find the circuit elements.
Fig. Q.4
rrr 5-3.6
The three-element series circuit contains one inductance
L 5 0.02 H. The applied voltage and resulting currents are shown
on the phasor diagram in Fig. Q.6. If v 5 500 rad/sec, what are the
two circuit elements?
rrr 5-3.7
Fig. Q.6
For the circuit shown in Fig. Q.7, the voltage across the inductor is
vL 5 15 sin 200 t. Find the total voltage and the angle by which the current lags the total voltage.
Fig. Q.7
195
Complex Impedance
Frequently Asked Questions linked to LO 3
r 5-3.1
A coil having a resistance of 10 kW and inductance of 50 mH is connected to a 10-volt, 10 kHz
power supply. Calculate the impedance.
[AU April/May 2011]
rr 5-3.2 A two-element series circuit is connected across ac source e(t) = 200 2 sin(wt + 20º)V. The current
in the circuit is then found to be i(t) =10 2 cos(wt – 25º) A Determine the parameters of the circuit.
[JNTU Nov. 2012]
5.4
PARALLEL CIRCUITS
The complex number system simplifies the analysis of parallel ac circuits.
In series circuits, the current is the same in all parts of the series circuit. In
parallel ac circuits, the voltage is the same across each element.
LO 4
5.4.1 Parallel RC Circuits
The voltages for an RC series circuit can be expressed using complex numbers, where the resistive voltage is
the real part of the complex voltage and the capacitive voltage is the imaginary part. For parallel RC circuits,
the voltage is the same across each component. Here, the total current can be represented by a complex
number. The real part of the complex current expression is the resistive current; the capacitive branch current
is the imaginary part.
EXAMPLE 5.7
A signal generator supplies a sine wave of 20 V, 5 kHz to
the circuit shown in Fig. 5.18. Determine the total current
IT , the phase angle and total impedance in the circuit.
Solution Capacitive reactance
XC =
1
1
=
=159.2 V
2pfC 6.28×5×103 ×0.2×10−6
Fig. 5.18
Since the voltage across each element is the same as the applied voltage, we can solve for the two branch
currents.
∴ current in the resistance branch
IR =
VS 20
=
= 0.2 A
R 100
and current in the capacitive branch
IC =
VS
20
=
= 0.126 A
X C 159.2
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
196
Circuits and Networks
The total current is the vector sum of the two branch currents.
∴ total current IT 5 (IR 1 jIC) A 5 (0.2 1 j0.13) A
In polar form, IT 5 0.24 ∠33°
So the phase angle u between applied voltage and total current is 33°. It indicates that the total line current
is 0.24 A and leads the voltage by 33°. Solving for impedance, we get
Z=
VS
20 ∠0o
=
= 83.3 ∠− 33o V
IT 0.24 ∠33o
5.4.2 Parallel RL Circuits
In a parallel RL circuit, the inductive current is imaginary and lies on the –j axis. The current angle is
negative when the impedance angle is positive. Here also, the total current can be represented by a complex
number. The real part of the complex current expression is the resistive current; and inductive branch current
is the imaginary part.
EXAMPLE 5.8
A 50 V resistor is connected in parallel with an inductive
reactance of 30 V. A 20 V signal is applied to the circuit.
Find the total impedance and line current in the circuit
shown in Fig. 5.19.
Solution Since the voltage across each element is the same
as the applied voltage, current in the resistive branch,
VS 20 ∠0o
=
= 0.4 A
R 50 ∠0o
Current in the inductive branch
IR =
IL =
VS
20 ∠0o
=
= 0.66 ∠− 90o
X L 30 ∠90o
Total current is
IT 5 0.4 2 j0.66
In polar form,
IT 5 0.77 ∠258.8°
Here the current lags behind the voltage by 58.8°.
Total impedance Z = VS
IT
=
20 –0o
0.77 –- 58.8o
= 25.97 –58.8o V
Fig. 5.19
197
Complex Impedance
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 4
rrr 5-4.1
For the parallel circuit shown in Fig. Q.1, find the currents
and the total current and construct the phasor diagram.
rrr5-4.2
Determine the voltage across each element in the circuit
shown in Fig. Q.2. Convert the circuit into an equivalent
series form. Draw the voltage phasor diagram.
rrr 5-4.3
Fig. Q.1
For the circuit shown in Fig. Q.3, the applied voltage v 5 Vm
cos vt. Determine the current in each branch and obtain the total current in terms of the cosine
function.
Fig. Q.2
Fig. Q.3
rrr 5-4.4
In a parallel circuit having a resistance R 5 5 V and L 5 0.02 H, the applied voltage is v 5 100
sin (1000 t 1 50°) volts. Find the total current and the angle by which the current lags the applied
voltage.
rrr 5-4.5
In the parallel circuit shown in Fig. Q.5, the current in the inductor is five times greater than the
current in the capacitor. Find the element values.
rrr 5-4.6
For the circuit shown in Fig. Q.6, find the total current and the magnitude of the impedance.
Fig. Q.5
5.5
Fig. Q.6
COMPOUND CIRCUITS
In many cases, ac circuits to be analysed consist of a combination of series
and parallel impedances. Circuits of this type are known as series-parallel, or
compound circuits. Compound circuits can be simplified in the same manner as a
series-parallel dc circuit consisting of pure resistances.
LO 5
198
Circuits and Networks
EXAMPLE 5.9
Determine the equivalent impedance of Fig. 5.20.
Solution In the circuit, Z1 is in series with the parallel combination
of Z2 and Z3.
where
Z1 5 (5 1 j10) V
Fig. 5.20
Z2 5 (2 2 j4) V
Z3 5 (1 1 j3) V
The total impedance
ZT = Z1 +
Z 2 Z3
Z 2 + Z3
( 2 − j 4) (1 + j 3)
= (5 + j10) +
( 2 − j 4) + (1 + j 3)
4.47– - 63.4∞ ¥ 3.16– + 71.5∞
3 - j1
14.12–81∞
= (5 + j10) +
3 - j1
= (5 + j10) +
= (5 + j10) +
14.12–81∞
3.16– - 18∞
5 5 1 j10 1 4.46 ∠26.1°
5 5 1 j10 1 4 1 j1.96
5 9 1 j11.96
The equivalent circuit for the compound circuit shown in Fig. 5.20 is a series circuit containing 9 V of
resistance and 11.96 V of inductive reactance. In polar form,
Z 5 14.96 ∠53.03°
The phase angle between current and applied voltage is
u 5 53.03°
EXAMPLE 5.10
In the circuit of Fig. 5.21, determine the values
of the following (a) ZT (b) IT (c) u.
Fig. 5.21
Complex Impedance
199
Solution First, the inductive reactance is calculated.
XL 5 2p f L
5 2p 3 50 3 0.1 5 31.42 V
In Fig. 5.22, the 10 V resistance is in series with the parallel combination of 20 V and j31.42 V.
Fig. 5.22
∴ Z =10 + (20) ( j 31.42)
T
(20 + j 31.42)
=10 +
628.4 ∠90o
37.24 ∠57.52o
=10 +16.87 ∠32.48o
5 10 1 14.23 1 j9.06 5 24.23 1 j 9.06
In polar form, ZT 5 25.87 ∠20.5°
Here, the current lags behind the applied voltage by 20.5°.
Total current
IT =
=
VS
ZT
20
= 0.77 ∠− 20.5o
25.87 ∠20.5
The phase angle between voltage and current is
u 5 20.5°
o
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 5
rrr 5-5.1
Determine the total impedance Z T , the total current I T , phase angle u, voltage across inductor
L, and voltage across resistor R 3 in the circuit shown in Fig. Q.1.
Fig. Q.1
200
Circuits and Networks
rrr 5-5.2
For the circuit shown in Fig. Q.2, determine
the total current IT , phase angle u and voltage
across each element.
rrr 5-5.3
In the parallel circuit shown in Fig. Q.3, the
applied voltage is v 5 100 sin 5000 t V. Find
the currents in each branch and also the total
current in the circuit.
iT
Fig. Q.2
i1
i2
Fig. Q.3
rrr 5-5.4
Solve for current I using PSpice in the circuit shown in Fig. Q.4.
Fig. Q.4
rrr
5-5.5 Using PSpice, find i0 for the circuit shown in Fig. Q.5.
2i0
t
Fig. Q.5
rrr 5-5.6
Using PSpice, find Thevenin equivalent circuit of the following Fig. Q.6 from terminals a – b and
c – d.
201
Complex Impedance
∠
∠
Fig. Q.6
rr 5-5.7
With PSpice, obtain i0 using superposition principle (Fig. Q.7).
Fig. Q.7
Frequently Asked Questions linked to LO 5
r 5-5.1
For the circuit shown in Fig. Q.1, determine the total current IT, phase angle, and power factor.
[AU Nov./Dec. 2012]
rr 5-5.2 In the two-mesh network shown in Fig. Q.2, determine the
[JNTU Nov. 2012]
a) mesh current, b) power supplied by the source, and c) power dissipated in each resistor.
100 µF 10 W
2W
IT
50 V
100Hz
30 W
–j5 W –j3 W
3W
0.1 H
Fig. Q.1
100 V
1W
j4 W
Fig. Q.2
Additional Solved Problems
PROBLEM 5.1
Calculate the total current in the circuit in Fig. 5.23,
and determine the voltage across resistor VR, and across
capacitor VC.
Fig. 5.23
202
Circuits and Networks
Solution Total impedance in the circuit shown in Fig. 5.23 is
Z = 100 ×103 − j
1
100p× 0.01×10−6
Z = 333.65×103 −72.56° V
The current in the circuit is
V
50
I= =
Z 333.65×103 −72.56ο
= 0.15×10−3 72.56° A
The voltage across resistor
VR = IR = 0.15×10−3 ×100 ×103 72.56°
= 15 72.56° V
The voltage across capacitor
VC = IX C = 0.15×10−3 72.56°×
1 −90°
100p× 0.01×10−6
= 47.75 −17.44° V
PROBLEM 5.2
A signal generator supplies a sine wave of 10 V, 10
kHz, to the circuit shown in Fig. 5.24(a). Calculate
the total current in the circuit. Determine the phase
angle u for the circuit. If the total inductance in the
circuit is doubled, does u increase or decrease, and
by how many degrees?
Solution In the circuit shown in Fig. 5.24 (a),
the inductances in mH are converted into ohms as
shown in Fig. 5.24 (b).
Fig. 5.24(a)
The total impedance in the circuit
Z 5 100 1 ( j125.66 || j314.15) V
= 100 + j 89.85 = 134.43 41.94° V
The total current in the circuit
10
I=
−41.94° = 0.074 −41.94°
134.43
If the inductance value is doubled, the impedance
Z = 100 + j 179.7 = 205.65 60.90° V
Hence, the u increased by 19°.
Fig. 5.24 (b)
203
Complex Impedance
PROBLEM 5.3
For the circuit shown in Fig. 5.25, determine the voltage
across each element. Is the circuit predominantly
resistive or inductive? Find the current in each branch
and the total current.
The circuit in Fig. 5.25 is converted as
shown in Fig. 5.26.
Solution
Fig. 5.25
Fig. 5.26
Consider the node voltage V1 at the node 1.
By applying Kirchhoff’s current law,
V1 − 30 0°
V1
V
V1
+
+ 1 +
=0
100
314.1 90° 500 942.5 90°
From Eq. (5.1), we get the node voltage
V1 = 23.1 19.71° = 21.75 + j 7.8
The voltage across the 100 V resistor = 30 0°− V1
5 30 2 21.75 2 j 7.8
5 8.25 2 j 7.8
= 11.35 −43.4° V
The voltage across remaining branch element is
V1 = 23.1 19.71° V
V100 11.35 −43.4°
=
100
100
= 0.113 −43.4° A
The current through the 100 V resistance =
I100
The current through the 500 V resistance
I 500 =
V1
23.1 19.71°
= 0.046 19.71° A
=
500
500
The current through the 1 mH inductor =
23.1 19.71°
= 0.073 −70.29° A
314.1 90°
(5.1)
204
Circuits and Networks
The current through the 3 mH inductor =
23.1 19.71°
= 0.024 −70.29° A
942.5 90°
Since the total current lags, the circuit is predominantly inductive.
Total current I = 0.113 −43.4° A .
PROBLEM 5.4
For the circuit shown in Fig. 5.27, determine the value of frequency
of supply voltage when a 100 V, 50 A current is supplied to the
circuit.
Solution Magnitude of the total current 5 50 A
Total impedance ZT =
10( j vL)
10 + j vL
Total current in the circuit IT =
IT =
Fig. 5.27
V
ZT
100(10 + jvL)
j10vL
∴ IT = 10 − j
100
vL
100 
The magnitude of the current = (10)2 + 
 = 50
 vL 
2
From the above equation, we have
v 5 0.816 and f 5 0.125 Hz.
PROBLEM 5.5
A sine-wave generator supplies a signal of 100 V,
50 Hz to the circuit shown in Fig. 5.28. Find the
current in each branch, and total current. Determine
the voltage across each element.
The inductance and capacitance values
are converted into ohms as shown in Fig. 5.29.
The node voltage = 100 0°
The current in (3 1 j31.41) V is
Solution
I1 =
100 0°
= (0.3 − j 3.15) A
3 + j 31.41
Fig. 5.28
205
Complex Impedance
The current in (5 2 j31.83) V is
I2 =
100 0°
= (0.48 + j 3.1) A
5 − j 31.83
The current in (10 1 j150.73) V is
Fig. 5.29
I3 =
100 0°
= (0.044 − j 0.66) A
10 + j150.73
The total current I 5 I1 1 I2 1 I3
I 5 (0.8242j0.71) A
The voltage across the 3 V resistor V3 = 3(0.3 − j 3.15) = 9.51 −84.5° V
The voltage across the 5 V resistor V5 = 5(0.48 + j 3.1) = 15.49 81.1° V
The voltage across the 10 V resistor V10 = 10(0.044 − j 0.66) = 6.6 −86.2° V
The voltage across the 0.1 H inductance
V0.1 H = (0.3 − j 3.15) j (31.41) = 99.35 5.44° V
The voltage across the 100 mF capacitance
V100 mF = (0.48 + j 3.1) ×(− j 31.83) = 99.62 −8.8° V
The voltage across the 0.5 H inductance
V0.5 H = (0.044 − j 0.66) ( j157.81) = 103.84 3.8° V
The voltage across the 500 mF capacitance
V500 mF = (0.044 − j 0.66) (− j 6.37) = 4.21 −176.2° V
PROBLEM 5.6
Determine the impedance and phase angle in the
circuit shown in Fig. 5.30.
Solution
Capacitive reactance X C =
=
1
2pfC
Fig. 5.30
1
2p×50× 100×10−6
= 31.83 V
206
Circuits and Networks
Capacitive susceptance BC =
=
Conductance G =
1
XC
1
= 0.031 S
31.83
1 1
= = 0.02 S
R 50
Total admittance Y = G 2 + BC2
= (0.02)2 + (0.031)2
= 0.037 S
1
1
Total impedance Z = =
= 27.02 V
Y 0.037
Phase angle
 R 

u = tan−1 
 X C 
 50 
= tan−1 
 31.83 
u = 57.52°
PROBLEM 5.7
For the parallel circuit in Fig. 5.31, find the magnitude of current in each branch and the total current. What
is the phase angle between the applied voltage and total current?
Fig. 5.31
Solution First let us find the capacitive reactances
X C1 =
1
2pf C1
=
XC2
1
= 31.83 V
2p×50×100×10−6
1
1
=
=
2pfC2 2p×50×300×10−6
= 10.61 V
Complex Impedance
207
Here, the voltage across each element is the same as the applied voltage.
Current in the 100 mF capacitor, I C1 =
=
VS
X C1
10 ∠0o
31.83 ∠− 90o
VS
X C2
Current in the 300 mF capacitor, I C2 =
=
= 0.31∠90o A
10 ∠0o
10.61∠− 90
o
= 0.94 ∠90o A
Current in the 100 V resistor is I R1 =
VS 10
=
= 0.1 A
R1 100
Current in the 200 V resistor is I R2 =
VS
10
=
= 0.05 A
R2 200
Total current IT 5 IR 1 IR 1 j(IC 1 IC )
1
2
1
2
5 0.1 1 0.05 1 j(0.31 1 0.94) 5 1.26 ∠83.2° A
The circuit shown in Fig. 5.31 can be simplified into a single parallel RC circuit as shown in Fig. 5.32.
Fig. 5.32
In Fig. 5.31, the two resistances are in parallel and can be combined into a single resistance. Similarly, the
two capacitive reactances are in parallel and can be combined into a single capacitive reactance.
R=
XC =
R1 R2
= 66.67 V
R1 + R2
X C1 X C2
X C1 + X C2
= 7.96 V
Phase angle u between voltage and current is
 R 
 66.67 
 = tan−1 
u = tan−1 
= 83.19°
 7.96 
 X C 
Here, the current leads the applied voltage by 83.19°.
208
Circuits and Networks
PROBLEM 5.8
For the circuit shown in Fig. 5.33, determine the total impedance, total current, and phase angle.
Fig. 5.33
Solution First, we calculate the magnitudes of the capacitive reactances.
X C1 =
X C2 =
1
2p×50×100×10−6
1
2p×50×300×10−6
= 31.83 V
= 10.61 V
We find the impedance of the parallel portion by finding the admittance.
G2 =
1
1
=
= 0.02 S
R2 50
BC2 =
1
1
=
= 0.094 S
X C 2 10.61
Y2 = G22 + BC22 = (0.02)2 + (0.094)2 = 0.096 S
Z2 =
1
1
=
= 10.42 V
Y2 0.096
The phase angle associated with the parallel portion of the circuit
uP 5 tan–1 (R2/XC ) 5 tan–1(50/10.61) 5 78.02°
2
The series equivalent values for the parallel portion are
Req 5 Z2 cos uP 5 10.42 cos (78.02°) 5 2.16 V
XC(eq) 5 Z2 sin uP 5 10.42 sin (78.02°) 5 10.19 V
The total resistance
RT 5 R1 1 Req
5 (10 1 2.16) 5 12.16 V
XCT 5 XC1 1 XC(eq)
5 (31.83 1 10.19) 5 42.02 V
Complex Impedance
209
Total impedance
ZT = RT2 + X C2T
= (12.16)2 + ( 42.02)2 = 43.74 V
We can also find the total current by using Ohm’s law.
V
100
IT = S =
= 2.29 A
ZT 43.74
The phase angle
X 
u= tan−1  CT 
 RT 
 42.02 
= tan−1 
= 73.86°
 12.16 
PROBLEM 5.9
Determine the voltage across each element of the circuit shown in Fig. 5.34 and draw the voltage phasor
diagram.
Fig. 5.34
Solution
First we calculate XL and XL
1
2
XL 5 2p f L1 5 2p 3 50 3 0.5 5 157.08 V
1
XL 5 2p f L2 5 2p 3 50 3 1.0 5 314.16 V
2
Now we determine the impedance of each branch
Z1 = R12 + X L21 = (100) 2 + (157.08) 2 =186.2 V
Z 2 = R22 + X L22 = (330) 2 + (314.16) 2 = 455.63 V
The current in each branch
I1 =
and
VS
100
=
= 0.537 A
Z1 186.2
I2 =
VS
100
=
= 0.219 A
Z2 455.63
210
Circuits and Networks
The voltage across each element
VR 5 I1R1 5 0.537 3 100 5 53.7 V
1
VL 5 I1XL 5 0.537 3 157.08 5 84.35 V
1
1
VR 5 I2R2 5 0.219 3 330 5 72.27 V
2
VL 5 I2XL 5 0.219 3 314.16 5 68.8 V
2
2
The angles associated with each parallel branch are now determined.
 X L 
157.08 
u1 = tan−1  1  = tan−1 
 100 
 R1 
= 57.52°
 X L 
 314.16 
u2 = tan−1  2  = tan−1 
 330 

 R2 
= 43.59°
i.e. I1 lags behind VS by 57.52° and I2 lags behind VS by
43.59°.
Here, VR and I1 are in phase and therefore, lag behind VS by
1
57.52°.
VR and I2 are in phase, and therefore lag behind VS by
2
43.59°.
VL leads I1 by 90°, so its angle is 90° – 57.52° 5 32.48°.
1
VL leads I2 by 90°, so its angle is 90° – 43.59° 5 46.41°.
2
The phase relations are shown in Fig. 5.35.
Fig. 5.35
PROBLEM 5.10
In the series–parallel circuit shown in Fig. 5.36, the effective value of voltage across the parallel parts of the
circuits is 50 V. Determine the corresponding magnitude of V.
Fig. 5.36
Solution
Here, we can determine the current in each branch of the parallel part.
Current in the j3 V branch, I1 =
50
= 16.67 A
3
211
Complex Impedance
Current in (10 1 j30) V branch, I 2 =
50
= 1.58 A
31.62
Total current IT 5 16.67 1 1.58 5 18.25 A
Total impedance ZT = 8.5 ∠30°+
= 8.5 ∠30°+
3 ∠90°×(10 + j 30)
(10 + j 30) + 3 ∠90°
3 ∠90°×31.62 ∠71.57°
10 + j 33
= 7.36 + j 4.25 +
94.86 ∠161.57°
34.48 ∠73.14°
5 7.36 1 j4.25 1 2.75∠88.43°
5 7.36 1 j4.25 1 0.075 1 j 2.75
5 (7.435 1 j7) V
5 10.21 ∠43.27°
In polar form, total impedance is ZT 5 10.21 ∠43.27°
The magnitude of applied voltage V 5 I 3 ZT 5 18.25 3 10.21 5 186.33 V.
PROBLEM 5.11
For the series parallel circuit shown in Fig. 5.37,
determine (a) the total impedance between the
terminals a, b and state if it is inductive or capacitive,
(b) the voltage across in the parallel branch, and (c)
the phase angle.
Solution Here, the parallel branch can be combined
into a single branch
ZP 5 (3 1 j4) || (3 1 j4) 5 (1.5 1 j2) V
Total impedance ZT 5 1 1 j2 1 1.5 1 j2 5 (2.5 1 j4) V
Hence, the total impedance in the circuit is inductive
Total current in the circuit
IT =
VS 10 + j 20
=
ZT 2.5 + j 4
=
22.36 ∠63.43°
4.72 ∠ 57.99°
Fig. 5.37
212
∴
Circuits and Networks
IT = 4.74 ∠5.44° A
i.e. the current lags behind the voltage by 57.99°.
Phase angle u 5 57.99°
Voltage across in the parallel branch
VP 5 (1.5 1 j2) 4.74 ∠5.44°
5 2.5 3 4.74 ∠(5.44° 1 53.13°)
5 11.85 ∠58.57° V
PROBLEM 5.12
In the series parallel circuit shown in Fig. 5.38, the two
parallel branches A and B are in series with C. The
impedances are ZA 5 10 1 j8, ZB 5 9 – j6, ZC 5 3 1 j2
and the voltage across the circuit is (100 1 j0) V.
Find the currents IA, IB and the phase angle between them.
Solution Total parallel branch impedance,
ZP =
Z A ZB
Z A + ZB
=
(10 + j8) (9 − j 6)
19 + j 2
=
12.8 ∠38.66°×10.82 ∠− 33.7 °
= 7.25 ∠−1.04°
19..1∠6°
In rectangular form,
Total parallel impedance ZP 5 7.25 – j0.13
This parallel impedance is in series with ZC
Total impedance in the circuit
ZT 5 ZP 1 ZC 5 7.25 2 j0.13 1 3 1 j2 5 (10.25 1 j1.87) V
Total current
IT =
=
VS
ZT
100 ∠0°
(100 + j 0)
=
4°
(10.25 + j1.87) 10.42 ∠10.34
= 9.6 ∠−10.34°
The current lags behind the applied voltage by 10.34°.
Fig. 5.38
Complex Impedance
213
Current in the branch A is
ZB
I A = IT
Z A + ZB
= 9.6 ∠−10.34°×
=
( 9 − j 6)
19 + j 2
9.6 ∠−10.34°×10.82 ∠− 33.7°
19.1∠6°
= 5.44 ∠− 50.04° A
Current in the branch B is IB
ZA
I B = IT ×
Z A + ZB
10 + j8
= 9.6 ∠−10.34°×
19 + j 2
=
9.6 ∠−10.34°×12.8 ∠38.66°
19.1∠6°
= 6.43 ∠22.32° A
The angle between IA and IB,
u 5 (50.04° 1 22.32°) 5 72.36°
PROBLEM 5.13
A series circuit of two pure elements has the following applied voltage and resulting current.
V = 15 cos (200 t – 30°) volts
I = 8.5 cos (200 t + 15) volts
Find the elements comprising the circuit.
Solution By inspection, the current leads the voltage by 30° 1 15° 5 45°. Hence, the circuit must contain
resistance and capacitance.
1
vCR
1
1=
, ∴
vCR
tan 45°=
1
=R
vC
 1 2
Vm
= R2 + R2 = 2 R
= R 2 + 
 vC 
Im
∴
R=
15
8.5× 2
=1.248 V
1
=1.248 V
vC
1
and C =
= 4×10−3 F
200×1.248
214
Circuits and Networks
PROBLEM 5.14
A resistor having a resistance of R 5 10 V and an unknown capacitor are in series. The voltage across
the resistor is VR 5 50 sin (1000 t 1 45°) volts. If the current leads the applied voltage by 60° what is the
unknown capacitance C?
Solution Here, the current leads the applied voltage by 60°.
1
tan 60°=
vCR
Since
R 5 10 V
v 5 1000 radians
tan 60° =
C=
1
vCR
1
= 57.7 mF
tan 60×1000×10
PROBLEM 5.15
A series circuit consists of two pure elements has the following current and voltage.
v = 100 sin (2000 t + 50°) V
i = 20 cos (2000 t + 20°) A
Find the elements in the circuit.
Solution We can write i 5 20 sin (2000 t 1 20° 1 90°)
Since cos u 5 sin (u 1 90°)
Current i 5 20 sin (2000 t 1 110°) A
The current leads the voltage by 110° – 50° 5 60°
and the circuit must consist of resistance and capacitance.
1
tan u =
vCR
1
= R tan 60°=1.73 R
vC
 1 2 100
Vm
=
= R2 + 
 vC 
20
Im
R 1+ (1.73)2 =
100
20
R (1.99) = 5
R = 2.5 V
and C =
1
=115.6 mF
v (1.73 R )
Complex Impedance
215
PROBLEM 5.16
A two-branch parallel circuit with one branch of R 5 100 V and a single unknown element in the other
branch has the following applied voltage and total current.
n 5 2000 cos (1000 t 1 45°) V
iT 5 45 sin (1000 t 1 135°) A
Find the unknown element.
Solution Here, the voltage applied is same for both elements.
Current passing through resistor is iR =
v
R
∴ iR 5 20 cos (1000 t 1 45°)
Total current iT 5 iR 1 iX
where ix is the current in unknown element.
ix 5 iT – iR
5 45 sin (1000 t 1 135°) – 20 cos (1000 t 1 45°)
5 45 sin (1000 t 1 135°) – 20 sin (1000 t 1 135°)
Current passing through the unknown element.
ix 5 25 sin (1000 t 1 135°)
Since the current and voltage are in phase, the element is a resistor.
And the value of the resistor
R=
v 2000
=
= 80 V
iX
25
PROBLEM 5.17
Find the total current to the parallel circuit with L 5 0.05 H and C 5 0.667 mF with an applied voltage of
v 5 200 sin 5000 t V.
1
Solution Current in the inductor iL = ∫ vdt
L
∴
iL =
=
1
200 sin 5000 t
0.05 ∫
−200 cos 5000 t
0.05×5000
iL =− 0.8 cos 5000 t
Current in the capacitor iC = C
∴
iC = 0.667 ¥ 10-6
dv
dt
d
(200 sin 5000t )
dt
216
Circuits and Networks
iC 5 0.667 cos 5000 t
Total current iT 5 iL 1 iC
5 0.667 cos 5000 t – 0.8 cos 5000 t
5 20.133 cos (5000 t)
Total current iT 5 0.133 sin (5000 t 2 90°) A
PSpice Problems
PROBLEM 5.1
For the parallel circuit shown in Fig. 5.39, find the magnitude of current in each branch and the total current.
What is the phase angle between the applied voltage and total current.
Fig. 5.39
* AC ANALYSIS
VS 1 0 AC 10 V0
C1 1 0 100 U
R1 1 0 100
C2 1 0 300 U
R2 1 0 200
.AC LIN 1 50 100
.PRINT AC IM(VS) IP(VS) IM(C1) IP(C1) IM(C2) IP(C2) IM(R1) 1 IP(R1) IM(R2) IP(R2)
.END
OUTPUT
**** AC ANALYSIS TEMPERATURE 5 27.000 DEG C
****************************************************************
FREQ
IM(VS)
IP(VS)
IM(C1)
IP(C1)
IM(C2)
5.000E101 1.266E100 – 9.681E101 3.142E – 01 9.000E101 9.425E – 01
FREQ
IP(C2)
IM(R1)
IP(R1)
IM(R2)
IP(R2)
5.000E101 9.000E101 1.000E – 01 0.000E100 5.000E – 02 0.000E100
Result
Frequency 5 50 Hz;
Complex Impedance
217
Total current It 5 I(VS) 5 1.266∠– 96.81 A
Ic1 5 0.3142∠90 A
Ic2 5 0.9425∠90 A
IR1 5 0.1∠0 A
IR2 5 0.05∠0 A
PROBLEM 5.2
Determine voltage across each element of circuit shown in Fig. 5.40.
Fig. 5.40
* AC ANALYSIS
VS 1 0 AC 100V 0
R1 1 2 100
L1 2 0 0.5
R2 1 3 330 OHM
L2 3 0 1H
.AC LIN 1 50 100
.PRINT AC V(R1) VP(R1) VM(L1) VP(L1) 1 V(R2) VP(R2) V(L2) VP(L2)
.END
**** AC ANALYSIS TEMPERATURE 5 27.000 DEG C
****************************************************************
FREQ
V(R1)
VP(R1)
VM(L1)
VP(L1)
V(R2)
5.000E101 5.370E101 25.752E101 8.436E101 3.248E101 7.243E101
FREQ
VP(R2)
V(L2)
VP(L2)
5.000E101
– 4.359E101
6.895E101
4.641E101
Result
VR1 5 53.7 ∠257.32 V
VL1 5 84.36 ∠32.48 V
VR2 5 72.43 ∠243.59 V
VL2 5 68.95 ∠46.41 V
PROBLEM 5.3
For the series–parallel circuit shown in Fig. 5.41 determine (a) total impedance between a, b, (b) voltage
across each parallel branch, and (c) phase angle.
218
Circuits and Networks
Fig. 5.41
* AC ANALYSIS OF SERIES PARALLEL CIRCUIT
VS
1
0
AC 22.36 63.435
R1
1
2
1
L1
2
3
6.366 M
R2
3
4
3
L2
4
0
12.732 M
R3
3
5
3
L3
5
0
12.732 M
.AC LIN
1
50
100
.PRINT AC IM(R1) IP(R1) VM(3,0) VP(3,0)
.END
OUTPUT
**** AC ANALYSIS TEMPERATURE 5 27.000 DEG C
****************************************************************
FREQ
IM(R1)
IP(R1)
VM(3,0)
VP(3,0)
5.000E 1 01 4.740E 1 00 5.441E 1 00
1.185E 1 01
5.857E 1 01
Result
(a) Total impedance between a and b : ZT 5 VS/IT 5 10 1 j20/11.85∠58.57° 5 4.72 ∠57.99°
(b) Voltage across each parallel branch 5 11.85 ∠58.77º
(c) Phase angle 5 63.435° – 5.441° 5 57.994°
Answers to Practice Problems
5-2.1 C 5 19.2 mF
5-3.1 157.4 ∠– 17.6°; 17.6° lead, 0.635 A
5-3.2 55.85 ∠257.5°; 57.5°
5-3.4 R 5 2 V; C 5 2.12 3 1023 F
5-3.5 R 5 5.412 V; C 5 160 3 1026 F
5-3.6 L150.02 H; L250.04 H
5-3.7 VT 5

vL 
R2 + (vL)2 lm sin vt + tan−1


R 
Phasor diagram
VT
Complex Impedance
u 5 tan−1
219
vL
, where v 5 200 rad/s
R
I1 = 10 53 1° A; I 2 = 5 0° A;
IT = 13.6 36° A
5-4.1
5-4.5 L 5 6.67 mH; C 5 3.33 mF
5-4.6 iT 5 1.74 sin (100t 1 67.4°) A
u 5 67.4°; Z 5 115 V
5-5.1 944.2 V; 0.053 A; 3.67°; 16.3 V; 30.7 V
5-5.2 1.44 A; 7.05°; V100 mF 5 22.9 V;
V10 V 5 14.4 V
V30 V 5 38.93 V; V0.1H 5 38.93 V
Objective-Type Questions
r 5.1
r 5.2
r 5.3
r 5.4
rr 5.5
rr 5.6
rr 5.7
r 5.8
r 5.9
r 5.10
r 5.11
r 5.12
A 1 kHz sinusoidal voltage is applied to an RL circuit, what is the frequency of the resulting current?
(a) 1 kHz
(b) 0.1 kHz
(c) 100 kHz
(d) 2 kHz
A series RL circuit has a resistance of 33 kV, and an inductive reactance of 50 kV. What is its impedance and
phase angle?
(a) 56.58 W, 59.9°
(b) 59.9 kW, 56.58°
(c) 59.9 V, 56.58°
(d) 5.99 V, 56.58°
In a certain RL circuit, VR 5 2 V and VL 5 3 V. What is the magnitude of the total voltage?
(a) 2 V
(b) 3 V
(c) 5 V
(d) 3.61 V
When the frequency of applied voltage in a series RL circuit is increased what happens to the inductive
reactance?
(a) Decreases
(b) Remains the same
(c) Increases
(d) Becomes zero
In a certain parallel RL circuit, R 5 0 V, and XL 5 75 V. What is the admittance?
(a) 0.024 S
(b) 75 S
(c) 50 S
(d) 1.5 S
What is the phase angle between the inductor current and the applied voltage in a parallel RL circuit?
(a) 0°
(b) 45°
(c) 90°
(d) 30°
When the resistance in an RC circuit is greater than the capacitive reactance, the phase angle between the
applied voltage and the total current is closer to
(a) 90°
(b) 0°
(c) 45°
(d) 120°
A series RC circuit has a resistance of 33 kV, and a capacitive reactance of 50 kV. What is the value of the
impedance?
(a) 50 kV
(b) 33 kV
(c) 20 kV
(d) 59.91 V
In a certain series RC circuit, VR 5 4 V and VC 5 6 V. What is the magnitude of the total voltage?
(a) 7.2 V
(b) 4 V
(c) 6 V
(d) 52 V
When the frequency of the applied voltage in a series RC circuit is increased what happens to the capacitive
reactance?
(a) It increases
(b) It decreases
(c) It is zero
(d) Remains the same
In a certain parallel RC circuit, R 5 50 V and XC 5 75 V. What is Y ?
(a) 0.01 S
(b) 0.02 S
(c) 50 S
(d) 75 S
The admittance of an RC circuit is 0.0035 S, and the applied voltage is 6 V. What is the total current?
(a) 6 mA
(b) 20 mA
(c) 21 mA
(d) 5 mA
220
Circuits and Networks
rrr 5.13 What is the phase angle between the capacitor current and the applied voltage in a parallel RC circuit?
(a) 90°
(b) 0°
(c) 45°
(d) 180°
rrr 5.14 In a given series RLC circuit, XC is 150 V, and XL is 80 V, what is the total reactance? What is the type of
reactance?
(a) 70 V, inductive
(b) 70 V, capacitive
(c) 70 V, resistive
(d) 150 V, capacitive
rrr 5.15 In a certain series RLC circuit, VR 5 24 V, VL 5 15 V, and VC 5 45 V. What is the source voltage?
(a) 38.42 V
(b) 45 V
(c) 15 V
(d) 24 V
rrr 5.16 When R 5 10 V, XC 5 18 V and XL 5 12 V, the current
(a) leads the applied voltage (b) lags behind the applied voltage
(c) is in phase with the voltage (d) is none of the above
rrr 5.17 A current i 5 A sin 500 t A passes through the circuit shown in Fig. 5.42. The total voltage applied will be
(a) B sin 500 t
(b) B sin (500 t – u°)
(c) B sin (500 t 1 u°)
(d) B cos (200 t 1 u°)
R
L
i
Fig. 5.42
rrr 5.18 A current of 100 mA through an inductive reactance of 100 V produces a voltage drop of
(a) 1 V
(b) 6.28 V
(c) 10 V
(d) 100 V
rrr 5.19 When a voltage v 5 100 sin 5000 t volts is applied to a series circuit of L 5 0.05 H and unknown capacitance,
the resulting current is i 5 2 sin (5000 t 1 90°) amperes. The value of capacitance is
(a) 66.7 pF
(b) 6.67 pF
(c) 0.667 mF
(d) 6.67 mF
rrr 5.20 A series circuit consists of two elements has the following current and applied voltage.
i 5 4 cos (2000 t 1 11.32°) A
v 5 200 sin (2000 t 1 50°) V
The circuit elements are
(a) resistance and capacitance
(b) capacitance and inductance
(c) inductance and resistance
(d) both resistances
rrr 5.21 A pure capacitor of C 5 35 mF is in parallel with another single circuit element. The applied voltage and
resulting current are
v 5 150 sin 300 t V
i 5 16.5 sin (3000 t 1 72.4°) A
The other element is
(a) capacitor of 30 mF
(b) inductor of 30 mH
(c) resistor of 30 V
(d) none of the above
For interactive quiz with answers,
scan the QR code given here
OR
visit
http://qrcode.flipick.com/index.php/263
6
LEARNING OBJECTIVES
6.1
INSTANTANEOUS POWER
In a purely resistive circuit, all the energy delivered by the source is dissipated
LO 1
in the form of heat by the resistance. In a purely reactive(inductive or capacitive)
circuit, all the energy delivered by the source is stored by the inductor or capacitor
in its magnetic or electric field during a portion of the voltage cycle, and then is
returned to the source during another portion of the cycle, so that no net energy is transferred. When there is
complex impedance in a circuit, a part of the energy is alternately stored and returned by the reactive part,
and part of it is dissipated by the resistance. The amount of energy dissipated is determined by the relative
values of resistance and reactance.
Consider a circuit having complex impedance. Let v (t) 5 Vm cos vt be the voltage applied to the circuit
and let i(t) 5 Im cos (vt 1 u) be the corresponding current flowing through the circuit. Then the power at any
instant of time is
P(t) 5 v (t) i (t)
5 Vm cos vt Im cos (vt 1 u)
(6.1)
From Eq. (6.1), we get
Fig. 6.1
Vm I m
(6.2)
[ cos ( 2 vt + u) + cos u]
2
Equation (6.2) represents instantaneous power. It
consists of two parts. One is a fixed part, and the other
is time-varying which has a frequency twice that of the
voltage or current waveforms. The voltage, current and
power waveforms are shown in Figs. 6.1 and 6.2.
P (t ) =
222
Circuits and Networks
Here, the negative portion (hatched) of the power
cycle represents the power returned to the source.
Figure 6.2 shows that the instantaneous power is
negative whenever the voltage and current are of
opposite sign. In Fig. 6.2, the positive portion of the
power is greater than the negative portion of the power;
hence, the average power is always positive, which is
almost equal to the constant part of the instantaneous
Fig. 6.2
power (Eq. 6.2). The positive portion of the power
cycle varies with the phase angle between the voltage
and current waveforms. If the circuit is pure resistive, the phase angle between voltage and current is zero; then
there is no negative cycle in the P (t) curve. Hence, all the
power delivered by the source is completely dissipated in
the resistance.
If u becomes zero in Eq. (6.1), we get
P(t) 5 v (t) i (t)
5 Vm Im cos2 vt
=
Vm I m
(1+ cos 2 vt )
2
(6.3)
The waveform for Eq. (6.3), is shown in Fig. 6.3, where
the power wave has a frequency twice that of the voltage or
current. Here, the average value of power is Vm Im /2.
When phase angle u is increased, the negative portion
of the power cycle increases and lesser power is dissipated.
When u becomes p/2, the positive and negative portions
of the power cycle are equal. At this instant, the power
dissipated in the circuit is zero, i.e. the power delivered to
the load is returned to the source.
Fig. 6.3
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 1*
rrr 6.1.1
Using PSpice, find the instantaneous power on each of the elements in the circuit of Fig. Q.1.
Fig. Q.1
*Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
223
Power and Power Factor
6.2
AVERAGE POWER
To find the average value of any power function, we have to take a particular time
interval from t1 to t2; by integrating the function from t1 to t2 and dividing the
result by the time interval t2 – t1, we get the average power.
LO 2
t2
1
Average power P =
P (t ) dt
t2 − t1 ∫t
(6.4)
1
In general, the average value over one cycle is
T
1
P (t ) dt
T ∫0
By integrating the instantaneous power P(t) in Eq. (6.5) over one cycle, we get average power
T

1 
Vm I m
Pav = ∫ 
cos (2vt + u) + cos u] dt 
[


T 

 2
Pav =
1
=
T
0
T
∫
0
Vm I m
1
[ cos (2vt + u) ] dt +
T
2
T
∫
Vm I m
2
cos u dt
(6.5)
(6.6)
0
In Eq. (6.6), the first term becomes zero, and the second term remains. The average power is therefore
Vm I m
Pav =
cosu W
(6.7)
2
We can write Eq. (6.7) as
 V  I 
Pav =  m  m  cosu
(6.8)
 2  2 
In Eq. (6.8), Vm 2 and I m
∴
Pav 5 Veff Ieff cos u
2 are the effective values of both voltage and current.
To get average power, we have to take the product of the effective values of both voltage and current
multiplied by cosine of the phase angle between voltage and the current.
If we consider a purely resistive circuit, the phase angle between voltage and current is zero. Hence, the
average power is
1
1
Pav = Vm I m = I m2 R
2
2
If we consider a purely reactive circuit (i.e. purely capacitive or purely inductive), the phase angle between
voltage and current is 90°. Hence, the average power is zero or Pav 5 0.
If the circuit contains complex impedance, the average power is the power dissipated in the resistive part
only.
EXAMPLE 6.1
A voltage of v(t) 5 100 sin vt is applied to a circuit. The current flowing through the circuit is
i(t) 5 15 sin (vt – 30°). Determine the average power delivered to the circuit.
Solution The phase angle between voltage and current is 30°.
224
Circuits and Networks
Effective value of the voltage Veff =
Effective value of the current I eff =
100
2
15
2
Average power Pav 5 Veff Ieff cos u
=
100
×
15
cos 30°
2
2
100×15
=
×0.866 = 649.5 W
2
EXAMPLE 6.2
Determine the average power delivered to the circuit consisting of an impedance Z 5 5 1 j8 when the
current flowing through the circuit is I 5 5  30°.
Solution The average power is the power dissipated in the resistive part only.
or
Current
∴
I m2
R
2
Im 5 5 A
Pav =
Pav =
52
×5 = 62.5 W
2
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 2
rrr6-2.1
Find the average power dissipated by the 500 V resistor shown in Fig. Q.1.
Fig. Q.1
rrr6-2.2
Find the power delivered by current source shown in Fig. Q.2.
Fig. Q.2
rrr 6-2.3
Using PSpice, find the average power absorbed by the 10 V resistor (Fig. Q.3).
225
Power and Power Factor
Fig. Q.3
6.3
APPARENT POWER AND POWER FACTOR
The power factor is useful in determining useful power (true power) transferred to
a load. The highest power factor is 1, which indicates that the current to a load is in
phase with the voltage across it (i.e. in the case of resistive load). When the power
factor is 0, the current to a load is 90° out of phase with the voltage (i.e. in case of
reactive load).
Consider the following equation:
Vm I m
Pav =
cosu W
2
In terms of effective values,
V I
Pav = m m cosu
2 2
5 Veff Ieff cos u W
LO 3
(6.9)
(6.10)
The average power is expressed in watts. It means the useful power transferred from the source to the
load, which is also called true power. If we consider a dc source applied to the network, true power is given
by the product of the voltage and the current. In case of sinusoidal voltage applied to the circuit, the product
of voltage and current is not the true power or average power. This product is called apparent power. The
apparent power is expressed in volt amperes, or simply VA.
∴
Apparent power 5 Veff Ieff
In Eq. (6.10), the average power depends on the value of cos u; this is called the power factor of the circuit.
Pav
∴
Power factor ( pf ) = cosu =
Veff I eff
Therefore, power factor is defined as the ratio of average power to the apparent power, whereas apparent
power is the product of the effective values of the current and the voltage. Power factor is also defined as the
factor with which the volt amperes are to be multiplied to get true power in the circuit.
In the case of sinusoidal sources, the power factor is the cosine of the phase angle between voltage and
current
pf 5 cos u
As the phase angle between voltage and total current increases, the power factor decreases. The smaller
the power factor, the smaller the power dissipation. The power factor varies from 0 to 1. For purely resistive
circuits, the phase angle between voltage and current is zero, and hence the power factor is unity. For
purely reactive circuits, the phase angle between voltage and current is 90°, and hence the power factor is
zero. In an RC circuit, the power factor is referred to as leading power factor because the current leads the
voltage. In an RL circuit, the power factor is referred to as lagging power factor because the current lags
behind the voltage.
226
Circuits and Networks
EXAMPLE 6.3
A sinusoidal voltage v 5 50 sin vt is applied to a series RL circuit. The current in the circuit is given by
i 5 25 sin (vt 2 53°). Determine (a) apparent power, (b) power factor, and (c) average power.
Solution (a) Apparent power P = Veff I eff
=
Vm
×
Im
2
2
50 × 25
=
= 625 VA
2
(b) Power factor 5 cos u
where u is the angle between voltage and current
u 5 53°
∴ Power factor 5 cos u 5 cos 53° 5 0.6
(c) Average power Pav 5 Veff Ieff cos u
5 625 3 0.6 5 375 W
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 3
rrr 6-3.1
For the circuit shown in Fig. Q.1, a voltage of 250 sin vt is applied. Determine the power factor
of the circuit, if the voltmeter readings are V1 5 100 V, V2 5 125 V, V3 5 150 V and the ammeter
reading is 5 A.
Fig. Q.1
A series RL circuit draws a current of i(t) 5 8 sin (50t 1 45°) from the source. Determine the
circuit constants, if the power delivered by the source is 100 W and there is a lagging power factor
of 0.707.
rrr 6-3.3 The current in a circuit lags the voltage by 30°. If the input power be 400 W and the supply voltage
be v 5 100 sin (377t 1 10°), find the complex power in voltamperes.
rrr 6-3.4 For the circuit shown in Fig. Q.4, determine the power dissipated and the power factor of the circuit.
rrr 6-3.2
Fig. Q.4
227
Power and Power Factor
rrr 6-3.5
For the circuit shown in Fig. Q.5, determine the power factor and the power dissipated in the
circuit.
Fig. Q.5
In the circuit shown in Fig. Q.6, the total effective
current is 30 amperes. Determine the three powers.
rrr 6-3.7 In the parallel circuit shown in Fig. Q.7, the total power
is 1100 watts. Find the power in each resistor and the
reading on the ammeter.
rrr 6-3.8 For the circuit shown in Fig. Q.8, determine the true
power, reactive power, and apparent power in each
branch. What is the power factor of the total circuit?
rrr6-3.6
Fig. Q.7
Fig. Q.6
Fig. Q.8
Determine the value of the voltage source, and the power factor in the network shown in Fig. Q.9
if it delivers a power of 500 W to the circuit shown in Fig. Q.9. Also find the reactive power drawn
from the source.
rrr6-3.10 Find the power dissipated by the voltage source shown in Fig. Q.10.
rrr6-3.9
Fig. Q.9
rrr 6-3.11
For the circuit shown in Fig.
Q.11, find:
(a) real power dissipated by
each element
(b) the total apparant power
supplied by the circuit
(c) the power factor of the
circuit.
Fig. Q.10
Fig. Q.11
228
Circuits and Networks
Frequently Asked Questions linked to LO 3*
rrr 6-3.1
Write the effect of power factor in energy-consumption billing.
(a) Determine the currents in all the branches
(b) Calculate the power and power factor of the source
(c) Show that power delivered by the source is equal to
+
power consumed by the 2-ohm resistor.
100Ð0ºV
[AU May/June 2014]
rrr 6-3.3 Calculate the power factor if v(t) = Vm sin wt and i(t) = Im
sin (wt – 45º).
[AU April/May 2011]
rrr 6-3.4 Define power factor.
rrr 6-3.2
6.4
[AU May/June 2014]
j 2.5 W
2W
–j 1 W
Fig. Q.2
[JNTU Nov. 2012]
REACTIVE POWER
We know that the average power dissipated is
LO 4
Pav 5 Veff [Ieff cos u]
(6.11)
From the impedance triangle shown in Fig. 6.4,
cosu=
R
Z
(6.12)
and Veff 5 Ieff Z
(6.13)
If we substitute Eqs (6.12) and (6.13) in Eq. (6.11), we get

R
Pav = I eff Z  I eff 

Z 
5 I 2eff R watts
(6.14)
This gives the average power dissipated in a resistive circuit.
If we consider a circuit consisting of a pure inductor, the power in
the inductor
Pr 5 ivL
(6.15)
Fig. 6.4
=iL
di
dt
Consider
i 5 Im sin (vt 1 u)
Then
Pr 5 I 2m sin (vt 1 u) Lv cos (vt 1 u)
I m2
(vL) sin 2 (vt + u)
2
∴
Pr 5 I 2eff (vL) sin 2 (vt 1 u)
(6.16)
From the above equation, we can say that the average power delivered to the circuit is zero. This is
called reactive power. It is expressed in volt-amperes reactive (VAR).
Pr 5 I 2eff XL VAR
(6.17)
=
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
229
Power and Power Factor
From Fig. 6.4, we have
XL 5 Z sin u
Substituting Eq. (6.18) in Eq. (6.17), we get
Pr 5 I 2eff Z sin u
(6.18)
5 (Ieff Z)Ieff sin u
5 Veff Ieff sin u VAR
6.5
THE POWER TRIANGLE
A generalised impedance phase diagram is shown in Fig. 6.5. A phasor relation for
LO 5
power can also be represented by a similar diagram because of the fact that true power
2 , as shown in Fig. 6.5.
Pav and reactive power Pr differ from R and X by a factor I eff
2
The resultant power phasor I eff Z, represents the apparent power Pa.
At any instant in time, Pa is the total power that appears to be transferred between the source and reactive
circuit. Part of the apparent power is true power and part of it is reactive power.
∴
Pa 5 I 2eff Z
The power triangle is shown in Fig. 6.6.
From Fig. 6.6, we can write
Ptrue 5 Pa cos u
or average power Pav 5 Pa cos u
and reactive power Pr 5 Pa sin u
Fig. 6.5
Fig. 6.6
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 5
rrr 6-5.1
Determine the branch and total real and reactive powers in the parallel circuit shown in Fig. Q.1.
Use j notation.
Fig. Q.1
230
Circuits and Networks
Two impedances Z1 5 (4 1 j5) V and Z2 5 (8 1 j10) V are connected in parallel across a 230 V,
50 Hz supply. Find (a) total admittance, (b) current drawn from supply and power factor, and (c)
value of capacitance to be connected in parallel with the above admittances to raise the power factor
unity.
rr 6-5.3 A voltage of v(t) 5 100 sin 500t is applied across a series R-L-C circuit where R 5 10 V, L 5 0.05
H, and C 5 20 mF. Determine the power supplied by the source, the reactive power supplied by the
source, the reactive power of the capacitor, the reactive power of the inductor, and the power factor
of the circuit.
rr 6-5.4 For the circuit shown in Fig. Q.4, determine the power factor, active power, reactive power, and
apparent power.
r 6-5.2
Fig. Q.4
For the parallel circuit shown in Fig. Q.5, the total power
dissipated is 1000 W. Determine the apparent power, the
reactive power, and the power factor.
rrr6-5.6 For the circuit shown in Fig. Q.6, determine the power
factor, active power, reactive power and apparent power.
rrr6-5.5
Fig. Q.5
Fig. Q.6
Additional Solved Problems
PROBLEM 6.1
For the circuit shown in Fig. 6.7, a voltage v(t) is applied and the resulting
current in the circuit i(t) 5 15 sin (vt 1 30°) amperes. Determine the active
power, reactive power, power factor, and the apparent power.
Solution The voltage applied to the circuit
v(t ) = 250 sin (vt + 100°)
V = 250 100°
Fig. 6.7
231
Power and Power Factor
The current in the circuit
i(t ) = 15 sin (vt + 30°)
I = 15 30°
Phase angle u 5 100 2 30 5 70°
Power factor cos u 5 cos 70° 5 0.342
250 15
Active power = VI cos u =
× × 0.342
2
2
= 641.25 watts
250 15
× ×sin 70°
Reactive power = VI sin u =
2
2
= 1762.5 VAR
Apparent power = VI =
250
2
×
15
2
= 1875 VA
PROBLEM 6.2
Two impedances, Z1 = 10 −60° V and Z = 16 70° V are in series and pass an effective current of 5 A.
Determine the active power, reactive power, apparent power and power factor.
Z1 = 10 −60° and Z 2 = 16 70°
Solution
Total Impedance Z 5 Z1 1 Z2
= 10 −60° + 16 70° = (10.47 + j 6.37) V
R 10.47
=
= 0.855
Z 12.25
Active power 5 VI cos u 5 I 2R 5 (5)2 3 10.47 5 261.75 watts
The power factor cos u =
Apparent power 5 I 2Z 5 (5)2 3 12.25 5 306.25 VA
Reactive power 5 I 2XL 5 (5)2 3 6.37 5 159.25 VAR
PROBLEM 6.3
For the circuit shown in Fig. 6.8, determine the value of the impedance if
the source delivers a power of 200 W and there is a lagging power factor
of 0.707. Also find the apparent power.
Solution The lagging power factor
cos u =
R
= 0.707
Z
R 5 Z(0.707)
Active power 5 VI cos u 5 200 watts
where
V=
25
2
= 17.68
Fig. 6.8
232
Circuits and Networks
25
× I cosu = 200
2
I 5 16 A
The total impedance ZT =
V 17.68 25°
=
= 1.105 45°
16 −20°
I
ZT 5 0.78 1 j 0.78
The impedance Z 5 ZT 2 R 5 (0.28 1 j 0.78) V
Apparent power 5 VI 5 17.68 3 16 5 282.7 VA.
PROBLEM 6.4
In the parallel circuit shown in Fig. 6.9, the power in the
5 V resistor is 600 W and the total circuit takes 3000 VA at a
leading power factor of 0.707. Find the value of Impedance
Z.
Solution Power in the 5 V resistor
I52R 5 600 W
I52 3 5 5 600 W
Current in the 5 V resistor I5 5 10.95 A
I5 =
V 0°
A
5 + j5
The magnitude of the current I 5 =
∴
Fig. 6.9
V
A
52 + 52
V = 10.95× 50 = 77.42 volts
Apparent power Pap 5 3000 VA
Pap
3000
Total current IT =
=
= 38.75 A
V
77.42
IT = 38.75 45°A
I 5 = 10.95 −45°A
The current in Z 5 IZ 5 IT 2 I5
5 27.4 1 j 27.4 2(7.74 2 j 7.74)
= 19.66 + j 35.14 = 40.26 60.77°
The value of impedance = Z =
Z=
V 0°
IZ
77.42 0°
40.26 60.77°
The value of impedance Z = 1.923 −60.77° V
233
Power and Power Factor
PROBLEM 6.5
A voltage source v(t) 5 150 sin vt in series with 5 V resistance is supplying two loads in parallel,
Z A = 60 30 and Z B = 50 -25 . Find the average power delivered to ZA, the average power delivered to
ZB the average power dissipiated in the circuit, and the power factor of the circuit.
Solution The circuit shown in Fig. 6.10 indicates
the voltage source v(t) in series with the 5 V resistor
provides supply to two parallel impedances ZA and ZB.
where Z A = 60 30° V; Z B = 50 −25° V
Total impedance
Z Z
60 30°×50 −25°
ZT = 5 + A B = 5 +
Z A + ZB
60 30° + 50 −25°
= 35.709 − 0.112 j
ZT = 35.71 −0.179° V
Fig. 6.10
Power factor 5 cos(0.179) 5 0.999
150
= 2.97 0.179° A
The total current I =
2 ZT
The current in impedance ZA is
IA = I×
ZB
2.97 0.179°×50 −25°
=
(Z A + ZB )
60 30° + 50 −25°
I A = 1.52 −30.03°A
The current in impedance ZB is
IB = I ×
ZA
2.97 0.179°×60 30°
=
(Z A + ZB )
60 30° + 50 −25°
I B = 1.82 24.96° A
Average power delivered to ZA 5 IA2RA 5 120 watts
Average power delivered to ZB 5 IB2RB 5 150 watts
Average power delivered to circuit 5 I 2R 5 314.98 watts
PROBLEM 6.6
A sine wave of v(t) 5 200 sin 50 t is applied to a 10 V resistor in series with a coil. The reading of a voltmeter
across the resistor is 120 V and across the coil, 75 V. Calculate the power and reactive volt-amperes in the
coil and the power factor of the circuit.
Solution The rms value of the sine wave
200
V=
=141.4 V
2
234
Circuits and Networks
Voltage across the resistor, VR 5 120 V
Voltage across the coil, VL 5 75 V
∴
IR = 120 V
120
= 12 A
10
IX L = 5 V
The current resistor, I =
Since
75
= 6.25 V
12
R
Power factor, pf = cosu =
Z
where Z 5 10 1 j 6.25 5 11.8∠32°
∴
∴
XL =
cos u =
R
10
=
= 0.85
Z 11.8
True power Ptrue 5 I 2 R 5 (12)2 3 10 5 1440 W
Reactive power Pr 5 I 2 XL 5 (12)2 3 6.25 5 900 VAR
PROBLEM 6.7
For the circuit shown in Fig. 6.11, determine the true power,
reactive power, and apparent power in each branch. What is
the power factor of the total circuit?
Solution In the circuit shown in Fig. 6.11, we can calculate
Z1 and Z2.
Impedance Z1 =
100∠15°
= 2∠5° = (1.99 + j 0.174) V
50∠10°
Impedance Z 2 =
100∠15°
= 5∠−15° = ( 4.83 − j1.29) V
20∠30°
True power in branch Z1 is Pt1 5 I 21 R 5 (50)2 3 1.99 5 4975 W
Reactive power in branch Z1, Pr1 5 I 21 XL
5 (50)2 3 0.174 5 435 VAR
Apparent power in branch Z1, Pa1 5 I 21 Z1
5 (50)2 3 2
5 2500 3 2 5 5000 VA
True power in branch Z2, Pt2 5 I 22 R
5 (20)2 3 4.83 5 1932 W
Reactive power in branch Z2, Pr2 5 I 22 XC
5 (20)2 3 1.29 5 516 VAR
Fig. 6.11
Power and Power Factor
235
Apparent power in branch Z2, Pa2 5 I 22 Z2
5 (20)2 3 5 5 2000 VA
Z1 Z2
Z1 + Z2
2 ∠5o ×5×∠−15o
=
1.99 + j 0.174 + 4.83 − j1.29
10 ∠−10o
=
6.82 − j1.116
10 ∠−10o
=
=1.45 ∠− 0.71o
6.9 ∠− 9.29
The phase angle between voltage and current, u 5 0.71°
∴
Power factor pf 5 cos u
5 cos 0.71° 5 0.99 leading
Total impedance of the circuit, Z =
PROBLEM 6.8
A voltage of v(t) 5 141.4 sin vt is applied to the circuit shown in Fig. 6.12. The circuit dissipates 450 W at a
lagging power factor, when the voltmeter and ammeter readings are 100 V and 6 A, respectively. Calculate
the circuit constants.
Fig. 6.12
Solution The magnitude of the current passing through (10 1 jX2) V is
I56A
The magnitude of the voltage across the (10 1 jX2) ohms, V 5 100 V. The magnitude of impedance (10 1 jX2)
is V/I.
100
Hence
10 2 + X 22 =
=16.67 V
6
∴
X 2 = (16.67) 2 − (10) 2 =13.33 V
Total power dissipated in the circuit 5 VI cos u 5 450 W
141.4
∴
V=
=100 V
2
I =6 A
100 ×6 × cos u = 450
450
= 0.75
The power factor pf = cos u =
600
u = 41.4°
236
Circuits and Networks
The current lags behind the voltage by 41.4°
The current passing through the circuit, I = 6∠− 41.4°
The voltage across (10 1 j13.33) V, V = 6∠− 41.4°×16.66∠53.1°
= 100∠11.7°
The voltage across parallel branch, V1 = 100∠0°−100∠11.7°
5 100 – 97.9 – j20.27
5 (2.1 – j20.27)V 5 20.38 – 84.08°
The current in (–j20) branch, I 2 =
20.38∠− 84.08°
20∠− 90o
= 1.02∠ + 5.92°
The current in (R1 – jX1) branch, I1
5 6∠– 41.4° – 1.02 5.92° 5 4.5 – j3.97 – 1.01 – j0.1
5 3.49 – j4.07 5 5.36 – 49.39°
V1 20.38∠− 84.08°
=
I1 5.36∠− 49.39°
5 3.8∠– 34.69° 5 (3.12 – j2.16) V
Z1 =
The impedance
Since R1 2 jX1 5 (3.12 – j2.16) V
R1 5 3.12 V
X1 5 2.16 V
PROBLEM 6.9
Determine the value of the voltage source and power
factor in the following network if it delivers a power of
100 W to the circuit shown in Fig. 6.13. Find also the
reactive power drawn from the source.
Solution Total impedance in the circuit,
Zeq
(2 + j 2)(− j 5)
= 5+
2 + j 2 − j5
10 − j10
14.14∠− 45°
= 5 + 3.93 ∠11.3°
=5+
2 − j3
3o
3.6∠− 56.3
= 5 + 3.85 + j 0.77 = 8.85 + j 0.77 = 8.88∠4.97°
=5+
Power delivered to the circuit, PT 5 I 2RT 5 100 W
∴
I 2 3 8.85 5 100
Current in the circuit, I =
100
= 3.36 A
8.85
Fig. 6.13
237
Power and Power Factor
Power factor pf = cos u =
=
Since
∴
R
Z
8.85
= 0.99
8.88
VI cos u 5 100 W
V 3 3.36 3 0.99 5 100
100
V=
= 30.06 V
3.36×0.99
The value of the voltage source, V 5 30.06 V
Reactive power Pr 5 VI sin u
5 30.06 3 3.36 3 sin (4.97°)
5 30.06 3 3.36 3 0.087
5 8.8 VAR
PROBLEM 6.10
For the circuit shown in Fig. 6.14, determine the
circuit constants when a voltage of 100 V is applied
to the circuit, and the total power absorbed is 600
W. The circuit constants are adjusted such that the
currents in the parallel branches are equal and
the voltage across the inductance is equal and in
quadrature with the voltage across the parallel
branch.
Fig. 6.14
Solution Since the voltages across the parallel branch and the inductance are in
quadrature, the total voltage becomes 100∠45° as shown in Fig. 6.15.
Total voltage is 100∠45° 5 V 1 j0 1 0 1 jV
From the above result, 70.7 1 j70.7 5 V 1 jV
∴ V 5 70.7
If we take current as the reference, then current passing through the circuit is
I∠0°. Total power absorbed by the circuit 5 VI cos u 5 600 W
or 100 3 I 3 cos 45° 5 600 W
∴ I 5 8.48 A
Hence, the inductance, X1 =
∴
V ∠90° 70.7 ∠90°
= 8.33 ∠90°
=
8.48
I ∠0°
X1 5 8.33 V
Current through the parallel branch R1 is I/2 5 4.24 A
Fig. 6.15
238
Circuits and Networks
Resistance, R1 =
70.7
V ∠0
=
= 16.67 V
I / 2 ∠0 4.24
Current through parallel branch R2 is I/2 5 4.24 A
Resistance is R2 =
70.7
=16.67 V
4.24
PROBLEM 6.11
Determine the average power delivered by the 500 0°
voltage source in Fig. 6.16 and also the dependent source.
Solution The current I can be determined by using
Kirchhoff’s voltage law.
I=
500 ∠0°− 3v4
7+4
Fig. 6.16
where v4 5 4I
I=
500∠0° 12 I
−
11
11
I = 21.73∠0°
500×21.73
= 5.432 kW
2
3v × I 3×4 I × I
=
= 2.833 kW
Power delivered by the dependent voltage source = 4
2
2
Power delivered by the 500 0° voltage source =
PROBLEM 6.12
Find the average power delivered by the dependent voltage source in the circuit shown in Fig. 6.17.
Fig. 6.17
Solution
The circuit is redrawn as shown in Fig. 6.18.
Fig. 6.18
239
Power and Power Factor
Assume current I1 flowing in the circuit.
The current I1 can be determined by using Kirchhoff’s voltage law.
I1 =
I1 −
100∠20° + 10 ×5 I1
5 + j4
50 I1
100∠20°
=
5 + j4
5 + j4
I1 = 2.213∠−154.9
9°
Average power delivered by the dependent source
Vm I m
= cos u
2
10V5 I1
=
cos u
2
=
50 ×(2.213)
2
=
2
= 122.43 W
PROBLEM 6.13
For the circuit shown in Fig. 6.19, find the average power
delivered by the voltage source.
Solution Applying Kirchhoff’s current law at the
node,
V − 50 Vx
V −100∠0°
V
+
+
=0
2
1 + j3
− j4
Vx =
Fig. 6.19
V
volts
1 + j3
Substituting in the above equation, we get
V −100∠0°
V
V
50 V
+
+
−
=0
2
1 + j 3 − j 4 (1 + j 3)(− j 4)
V = 14.705 ∠157.5°
I=
V −100 ∠0o 14.705 ∠157.5°−100 ∠0°
=
= 56.865 ∠177.18°
2
2
100×56.865 cos 177.18°
2
5 2.834 kW
Power delivered by the source =
240
Circuits and Networks
PROBLEM 6.14
For the circuit shown in Fig. 6.20, find the average
power delivered by the dependent current source.
Solution Applying Kirchhoff’s current law at the node,
V − 20∠0°
V
− 0.5V1 +
=0
10
20
Fig. 6.20
where V1 = 20∠0°−V
Substituting V1 in the above equation, we get
V = 18.46 ∠0°
V1 = 1.54 ∠0°
Average power delivered by the dependent source
Vm I m cos u 18.46×0.5×1.54
=
= 7.107 W
2
2
PSpice Problems
PROBLEM 6.1
Determine the power factor, true power, reactive power, and apparent
power using PSpice in the circuit shown in Fig. 6.21.
f 5 50 Hz
1
c=
=15.915 mF
2pf ×c
* PROGRAM FOR OBTAINING PF, S, P, AND Q
VS
1
0
AC 50 0
R1
1
2
100
C1
2
0
15.915 U
.AC LIN 1
50
100
.PRINT AC IM(R1)IP(R1)
.END
**** AC ANALYSIS TEMPERATURE 5 27.000 DEG C
****************************************************************
Fig. 6.21
241
Power and Power Factor
FREQ
5.000E 1 01
IM(R1)
2.236E – 01
IP(R1)
6.344E 1 01
Result
Power factor 5 COS(63.44) 5 0.4471 Lead
True power 5 VICOSf 5 50 3 0.2236 3 COS(63.44)
55W
Reactive power 5 VISINf 5 50 3 0.2236 3 SIN(63.44)
5 10 VAr
Apparent power 5 VI 5 50 3 0.2236 5 11.18 VA
PROBLEM 6.2
For the circuit shown in Fig. 6.22, determine the
average power delivered by the 50 ∠0° voltage source
and dependent source using PSpice.
Fig. 6.22
Vs =
500
= RMS Value
2
= 353.55 V
v = 40 rad / sec
v
= 6.366 Hz
2p
* PROGRAM TO FIND AVERAGE POWER
VS 1
0
AC 353.55 0
R1 1
2
4
R2 2
3
7
E3 0
1
2
3
.AC LIN 1 6.366 10
.PRINT AC I(VS), IP(VS), VM(R1), VP(R1)
.END
**** AC ANALYSIS TEMPERATURE 5 27.000 DEG C
****************************************************************
FREQ
I(VS)
IP(VS)
VM(R1)
VP(R1)
6.366E 1 00 1.537E 1 01 1.800E 1 02 6.149E 1 01
0.000E 1 00
Result It is a pure resistive circuit.
Power delivered by the 500∠0° source 5 VS 3 I 5 353.55 3 15.37
5 54.34 kW
Power delivered by dependent source 5 3V4 3 I 5 3 3 61.49 3 15.37
5 2.835 kW
⇒ f =
242
Circuits and Networks
Answers to Practice Problems
6-2.1 0.0812 mW
6-3.8 15.396 kW; 3944 VAR; 15.87 KVA; 0.97
6-2.2
– 0.114 W
PT real = 1800 w;
PT reactive = 600 VAR lagging
6-3.1
0.97
P10 real = 100 w;
P10 reactive = 0; P(8+j6) real = 800 w
6-3.2
3.12 V, 9.93 H
P(8+j6)reactive = 600 VAR
6-5.1
6-3.3 Pcomplex = 461.8 VA
6-5.2
486.5; 0.27
6-3.4
(ii) Current drawn = 53.825 A
Pactive = 2160 watts;
6-3.6
Preactive = 483 VAR leading
Power factor = 0.6238
Papparent = 2210 VA
6-3.7 P10 = 500 W;
(i) Total admittance = 0.234 −51.4° �
(iii) C = 5.79 × 1024 F
P3 = 600 W;
6-5.4
Ammeter reading = 19.25 −36° A
0.891; 1587.7 W; 806.2 VAR; 1781.9 VA
6-5.5 1136.36 VA; 529.6 VAR; 0.88
Objective-Type Questions
rrr 6.1
The phasor combination of resistive power and reactive power is called
(a) true power
rrr 6.2
(b) watts
(c) volt-amperes or watts (d) VAR
(c) combination of both (a) and (b)
(d) none of these
A power factor of ‘0’ indicates
(a) purely resistive element
(b) purely reactive element
rrr 6.5
(c) combination of both (a) and (b)
(d) none of the above
For a certain load, the true power is 100 W and the reactive power is 100 VAR. What is the apparent power?
(a) 200 VA
(b) 100 VA
(c) 141.4 VA
rrr 6.6
If a load is purely resistive and the true power is 5 W, what is the apparent power?
rrr 6.7
(a) 10 VA
True power is defined as
(a) VI cos u
rrr 6.8
(d) 120 VA
50 VA
(b) 5 VA
(c) 25 VA
(d)
(b) VI
(c) VI sin u
(d) none of these
In a certain series RC circuit, the true power is 2 W, and the reactive power is 3.5 VAR. What is the apparent
power?
(a) 3.5 VA
rrr 6.9
(d) average power
A power factor of ‘1’ indicates
(a) purely resistive circuit
(b) purely reactive circuit
rrr 6.4
(b) apparent power
Apparent power is expressed in
(a) volt-amperes
rrr 6.3
(c) reactive power
(b) 2 VA
If the phase angle u is 45°, what is the power factor?
(a) cos 45°
(b) sin 45°
(c) 4.03 VA
(d) 3 VA
(c) tan 45°
(d) none of these
rrr 6.10 To which component in an RC circuit is the power dissipation due?
(a) Capacitance
(b) Resistance
(c) Both
(d) None
243
Power and Power Factor
rrr 6.11 A two element series circuit with an instantaneous current I 5 4.24 sin (5000 t 1 45°) A has a power of 180
watts and a power factor of 0.8 lagging. The inductance of the circuit must have the value.
(a) 3 H
(b) 0.3 H
(c) 3 mH
(d) 0.3 mH
rrr 6.12 In the circuit shown in Fig. 6.23, if branch A takes 8 KVAR,
the power of the circuit will be
(a) 2 kW
(b) 4 kW
(c) 6 kW
(d) 8 kW
rrr 6.13 In the circuit shown in Fig. 6.24, the voltage across 30 V
resistor is 45volts. The reading of the ammeter A will be
(a) 10 A
(b) 19.4 A
Fig. 6.23
(c) 22.4 A
(d) 28 A
rrr6.14 In the circuit shown in Fig. 6.25, V 1 and V 2 are two
identical sources of 10∠90°. The power supplied by V1
is
(a) 6 W
(b) 8.8 W
(c) 11 W
(d) 16 W
Fig. 6.24
Fig. 6.25
For interactive quiz with answers,
scan the QR code given here
OR
visit
http://qrcode.flipick.com/index.php/264
7
LEARNING OBJECTIVES
7.1
MESH ANALYSIS
We have earlier discussed mesh analysis but have applied it only to resistive
LO 1
circuits. Some of the ac circuits presented in this chapter can also be solved
by using mesh analysis. In Chapter 2, the two basic techniques for writing
network equations for mesh analysis and node analysis were presented. These
concepts can also be used for sinusoidal steady-state condition. In the sinusoidal
steady-state analysis, we use voltage phasors, current phasors, impedances and
admittances to write branch equations, KVL and KCL equations. For ac circuits, the method of writing loop
equations is modified slightly. The voltages and currents in ac circuits change polarity at regular intervals.
At a given time, the instantaneous voltages are driving in either the positive or negative direction. If the
impedances are complex, the sum of their voltages is found by vector addition. We shall illustrate the method
of writing network mesh equations with the following example.
Consider the circuit shown in Fig. 7.1, containing a
voltage source and impedances.
The current in impedance Z1 is I1, and the current in
Z2, (assuming a positive direction downwards through the
impedance) is I1 2 I2. Similarly, the current in the impedance
Z3 is I2. By applying Kirchhoff’s voltage law for each loop,
we can get two equations. The voltage across any element
is the product of the phasor current in the element and the
Fig. 7.1
complex impedance.
245
Steady-State AC Analysis
Equation for the loop 1 is
I1Z1 1 (I1 2 I2)Z2 5 V1
(7.1)
Equation for the loop 2, which contains no source is
Z2(I2 2 I1) 1 Z3I2 5 0
By rearranging the above equations, the corresponding mesh current equations are
I1(Z1 1 Z2) 2 I2Z2 5 V1
(7.2)
(7.3)
– I1Z2 1 I2(Z2 1 Z3) 5 0
(7.4)
By solving the above equations, we can find out currents I1 and I2. In general, if we have M meshes, B
branches and N nodes including the reference node, we assume M branch currents and write M independent
equations; then the number of mesh currents is given by M 5 B 2 (N 2 1).
EXAMPLE 7.1
Write the mesh current equations in the circuit shown
in Fig. 7.2, and determine the currents.
Solution The equation for the loop 1 is
I1( j4) 1 6(I1 – I2) 5 5 ∠0°
The equation for the loop 2 is
(7.5)
Fig. 7.2
6(I2 2 I1) 1 ( j3)I2 1 (2)I2 5 0
(7.6)
By rearranging the above equations, the corresponding mesh current equations are
I1(6 1 j4) 2 6I2 5 5 ∠0°
26I1 1 (8 1 j3)I2 5 0
Solving the above equations, we have
 (8 + j 3) 
 I2
I1 = 
 6 
 (8 + j 3) (6 + j 4) 

 I 2 − 6 I 2 = 5 ∠0°


6

 (8 + j 3) (6 + j 4)
I2 
− 6 = 5 ∠0°


6
I2 [10.26 ∠ 54.2° 2 6 ∠0°] 5 5 ∠0°
I2 [(6 1 j8.32) 2 6] 5 5 ∠0°
I2 =
5 ∠0°
= 0.6 ∠−90°
8.32 ∠90°
I1 =
8.54 ∠20.5°
× 0.6 ∠−90°
6
(7.7)
(7.8)
246
Circuits and Networks
I1 5 0.855 ∠269.5°
Current in loop 1, I1 5 0.855 5 ∠269.5°
Current in loop 2, I2 5 0.6 ∠290°
7.2
MESH EQUATIONS BY INSPECTION
LO 1
In general, mesh equations can be written by observing any network. Consider the three-mesh network shown
in Fig. 7.3.
Fig. 7.3
The loop equations are
I1 Z1 1 Z2(I1 2 I2) 5 V1
Z2(I2 2 I1) 1 Z3 I2 1 Z4(I2 2 I3) 5 0
Z4(I3 2 I2) 1 Z5 I3 5 2V2
(7.9)
(7.10)
(7.11)
By rearranging the above equations, we have
(Z1 1 Z2)I1 2 Z2 I2 5 V1
2 Z2 I1 1 (Z2 1 Z3 1 Z4)I2 2 Z4I3 5 0
2 Z4 I2 1 (Z4 1 Z5)I3 5 2V2
(7.12)
(7.13)
(7.14)
In general, the above equations can be written as
Z11 I1 6 Z12 I2 1 Z13 I3 5 Va
(7.15)
6 Z21 I1 1 Z22 I2 6 Z23 I3 5 Vb
(7.16)
6 Z31 I1 6 Z32 I2 1 Z33 I3 5 Vc
(7.17)
If we compare the general equations with the circuit equations, we get the self-impedance of the loop 1
Z11 5 Z1 1 Z2
i.e. the sum of the impedances through which I1 passes. Similarly, Z22 5 (Z2 1 Z3 1 Z4), and Z33 5 (Z4 1 Z5)
are the self-impedances of loops 2 and 3. This is equal to the sum of the impedances in their respective loops,
through which I2 and I3 passes, respectively.
Z12 is the sum of the impedances common to loop currents I1 and I2. Similarly Z21 is the sum of the
impedances common to loop currents I2 and I1. In the circuit shown in Fig. 7.3, Z12 5 2Z2, and Z21 5 2Z2.
Here, the positive sign is used if both currents passing through the common impedance are in the same
direction; and the negative sign is used if the currents are in opposite directions. Similarly, Z13, Z23, Z31, Z32
are the sums of the impedances common to the mesh currents indicated in their subscripts. Va, Vb, and Vc are
247
Steady-State AC Analysis
sums of the voltages driving their respective loops. Positive sign is used, if the direction of the loop current
is the same as the direction of the source current. In Fig. 7.3, Vb 5 0 because no source is driving the loop 2.
Since the source, V2 drives against the loop current I3, Vc 5 2V2.
EXAMPLE 7.2
For the circuit shown in Fig. 7.4, write the mesh equations
using the inspection method.
Solution The general equations are
Z11 I1 6 Z12 I2 6 Z13 I3 5 Va
6 Z21 I1 1 Z22 I2 6 Z23 I3 5 Vb
(7.18)
(7.19)
6 Z31 I1 6 Z32 I2 1 Z33 I3 5 Vc
(7.20)
Fig. 7.4
Consider Eq. (7.18)
Z11 5 the self-impedance of the loop 1 5 (5 1 3 2 j4) V
Z12 5 the impedance common to both loops 1 and 2 5 25 V
The negative sign is used because the currents are in opposite directions.
Z13 5 0, because there is no common impedance between loop 1 and loop 3.
Va 5 0, because no source is driving the loop 1.
∴ Eq. (7.18) can be written as
(8 2 j4)I1 2 5I2 5 0
(7.21)
Now, consider Eq. (7.19).
Z21 5 25, the impedance common to loops 1 and 2.
Z22 5 (5 1 j5 2 j6), the self-impedance of the loop 2.
Z23 5 2 (2 j6), the impedance common to loops 2 and 3.
Vb 5 210 ∠30°, the source driving the loop 2.
The negative sign indicates that the source is driving against the loop current, I2.
Hence, Eq. (7.19) can be written as
2 5I1 1 (5 2 j1)I2 1 ( j6)I3 5 2 10 ∠30°
(7.22)
Consider Eq. (7.20).
Z31 5 0, there is no common impedance between loops 3 and 1
Z32 5 2 (2 j6), the impedance common to loops 2 and 3
Z33 5 (4 2 j6), the self impedance of the loop 3
Vb 5 20 ∠50°, the source driving the loop 3
The positive sign is used because the source is driving in the same direction as the loop current 3. Hence,
the equation can be written as
( j6)I2 1 (4 2 j6)I3 5 20 ∠50°
(7.23)
The three mesh equations are
(8 2 j4)I1 2 5I2 5 0
2 5I1 1 (5 2 j1)I2 1 ( j6)I3 5 210 ∠30°
( j6)I2 1 (4 2 j6)I3 5 20 ∠50
248
Circuits and Networks
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 1*
rrr 7-1.1
For the circuit shown in Fig. Q.1, determine the
value of the current Ix in the impedance Z 5 4 1 j5
between nodes a and b.
rrr 7-1.2
For the given network shown in Fig. Q.2, find the
current through (2 1 j3) V impedance using mesh
analysis.
Fig. Q.1
Fig. Q.2
rrr 7-1.3
For the circuit shown in Fig. Q.3, find the voltage across the dependent source branch by using mesh
analysis.
Fig. Q.3
rrr 7-1.4
By application of Thèvenin’s theorem, find the
current I in the circuit shown in Fig. Q.4. The
voltage source shown is sinusoidal having a 10 V
rms value and frequency is such that the inductor
has an impedance of 20 V magnitude and each
capacitor has an impedance of 10 V magnitude.
Fig. Q.4
Frequently Asked Questions linked to LO 1*
rrr 7-1.1
rrr 7-1.2
Define ‘mesh analysis’ of a circuit. [AU April/May 2011]
For the network shown in Fig. Q.2, obtain the current
ratio (I1/I3) using mesh analysis.
[AU April/May 2011] V
*For answers to Frequently Asked Questions, please visit the link http://highered.
mheducation.com/sites/9339219600
*Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
j2W
–j 4 W
5W
5W
j2W
1
I1
I2
Fig. Q.2
I3
249
Steady-State AC Analysis
rrr 7-1.3
rrr7-1.4
Using mesh analysis, determine the current the 1 W resistor of the network shown in Fig. Q.3. [BPUT 2008]
For the circuit shown in Fig. Q.4, determine the value of V2 such that the current through (3+j4) W
impedance is zero.
(AU Nov./Dec. 2012)
4W
6W
50V
10 V
4W
7.3
4W
4W
1W
V1 = 20 0°
I1
Fig. Q.3
3W
-j3 W I
2
j4 W
5W
-j5 W
I3
V2
Fig. Q.4
NODAL ANALYSIS
The
node-voltage
method can also be
used with networks
containing complex
impedances
and excited by
sinusoidal voltage sources. In general, in an
N-node network, we can choose any node as the
reference or datum node. In many circuits, this
Fig. 7.5
reference is most conveniently choosen as the
common terminal or ground terminal. Then it is possible to write (N 2 1) nodal equations using KCL. We
shall illustrate nodal analysis with the following example.
Consider the circuit shown in Fig. 7.5.
Let us take a and b as nodes, and c as reference node. Va is the voltage between nodes a and c. Vb is the voltage
between nodes b and c. Applying Kirchhoff’s current law at each
node, the unknowns Va and Vb are obtained.
In Fig. 7.6, node a is redrawn with all its branches,
assuming that all currents are leaving the node a.
LO 2
In Fig. 7.6, the sum of the currents leaving the node a is
zero.
∴ I1 1 I2 1 I3 5 0
Fig. 7.6
where
I1 =
(7.24)
V −Vb
Va −V1
V
, I 2 = a , I3 = a
Z3
Z1
Z2
Substituting I1, I2 and I3 in Eq. (7.24), we get
Va −V1 Va Va −Vb
+
+
=0
Z1
Z2
Z3
Fig. 7.7
(7.25)
Similarly, in Fig. 7.7, the node b is redrawn with all its
branches, assuming that all currents are leaving the node b.
250
Circuits and Networks
In Fig. 7.7, the sum of the currents leaving the node b is zero.
∴ I3 1 I4 1 I5 5 0
where
I3 =
(7.26)
Vb −Va
V
Vb
, I 4 = b , I5 =
Z5 + Z6
Z3
Z4
Substituting I3, I4 and I5 in Eq. (7.26)
we get
Vb
Vb −Va Vb
+
+
=0
Z3
Z 4 Z5 + Z 6
(7.27)
Rearranging Eqs 7.25 and 7.27, we get
1

 
 
 + 1 + 1 V −  1 V =  1 V
 a   b   1
 Z
Z 2 Z3 
 Z1 
 Z3 
1
1
1
1
1 
V = 0
+
− Va +  +
 Z3 Z 4 Z5 + Z6  b
 Z3 
(7.28)
(7.29)
From Eqs 7.28 and 7.29, we can find the unknown voltages Va and Vb.
EXAMPLE 7.3
In the network shown in Fig. 7.8, determine Va and Vb.
Fig. 7.8
Solution To obtain the voltage Va at a, consider the branch currents leaving the node a as shown in Fig. 7.9 (a).
In Fig. 7.9 (a), I1 =
V −Vb
Va −10 ∠0°
V
, I2 = a , I3 = a
− j6
3
j6
Fig. 7.9(a)
Since the sum of the currents leaving the node a is zero,
I1 1 I2 1 I3 5 0
251
Steady-State AC Analysis
Va −10 ∠0° Va
V −Vb
+
+ a
=0
− j6
j6
3
1

 − 1 + 1 V − 1 V = 10 ∠0°
 j 6 j 6 3  a 3 b
j6
∴
(7.30)
1
1
10 ∠0°
Va − Vb =
3
3
j6
(7.31)
To obtain the voltage Vb at b, consider the branch
currents leaving the node b as shown in Fig. 7.9 (b).
In Fig. 7.9 (b), I 3 =
Fig. 7.9(b)
V
Vb
Vb −Va
, I 4 = b , I5 =
3
( j 5 − j 4)
j4
Since the sum of the currents leaving the node b is zero,
I3 1 I4 1 I5 5 0
Vb −Va Vb Vb
+ + =0
3
j 4 j1
(7.32)
1 1
1
1
− Va +  + + Vb = 0
3
 3 j 4 j1
(7.33)
From Eqs (7.31) and (7.33), we can solve for Va and Vb,
0.33Va 2 0.33Vb 5 1.67 ∠290°
(7.34)
20.33Va 1 (0.33 2 0.25j 2 j)Vb 5 0
(7.35)
Adding Eqs (7.34) and (7.35), we get (21.25j)Vb 5 1.67 ∠290°
21.25 ∠90° Vb 5 1.67 ∠290°
Vb =
1.67 ∠−90°
−1.25 ∠90°
5 21.34 ∠2180°
Substituting Vb in Eq. (7.34), we get
0.33Va 2 (0.33) (21.34 ∠2180°) 5 1.67 ∠290°
Va =
1.67 ∠−90°
= −1.31 V
0.33
Va 5 5.22 ∠2104.5° V
Voltages Va and Vb are 5.22 ∠2104.5° V and 21.34 ∠2180° V respectively.
252
7.4
Circuits and Networks
NODAL EQUATIONS BY INSPECTION
LO 2
In general, nodal equations can also be written
by observing the network. Consider a four-node
network including a reference node as shown in
Fig. 7.10.
Consider nodes a, b, and c separately as shown
in Figs 7.11 (a), (b), and (c).
Assuming that all the currents are leaving the
nodes, the nodal equations at a, b and c are
I1 1 I2 1 I3 5 0
Fig. 7.10
I3 1 I4 1 I5 5 0
I5 1 I6 1 I7 5 0
Fig. 7.11
Va −V1 Va Va −Vb
+
+
=0
Z1
Z2
Z3
(7.36)
Vb −Va Vb Vb −Vc
+
+
=0
Z3
Z4
Z5
(7.37)
Vc −Vb Vc Vc −V2
+
+
=0
Z5
Z6
Z7
(7.38)
Steady-State AC Analysis
253
Rearranging the above equations, we get
 
1

 
 + 1 + 1 V −  1 V =  1 V


a
b
 Z 
 Z
 Z1  1
Z 2 Z3 
1
3
(7.39)
 −1


 
 V +  1 + 1 + 1 V −  1 V = 0


a
b
 Z
 Z  c
 Z3 
Z 4 Z5 
3
5
(7.40)
 −1


 
 V +  1 + 1 + 1 V =  1 V
 c   2
 Z  b  Z
Z6 Z 7 
 Z7 
5
5
(7.41)
In general, the above equations can be written as
YaaVa 1 YabVb 1 YacVc 5 I1
YbaVa 1 YbbVb 1 YbcVc 5 I2
YcaVa 1 YcbVb 1 YccVc 5 I3
If we compare the general equations with the circuit equations, the self-admittance at the node a is
1
1
1
Yaa = +
+
Z1 Z 2 Z3
which is the sum of the admittances connected to the node a.
1
1
1
1
1
1
Similarly, Ybb =
+
+ , and Ycc =
+
+
Z3 Z 4 Z5
Z5 Z6 Z 7
are the self admittances at nodes b and c, respectively. Yab is the mutual admittance between nodes a and b,
i.e. it is the sum of all the admittances connecting nodes a and b. Yab 5 21/Z3 has a negative sign. All the
mutual admittances have negative signs. Similarly, Yac, Yba, Ybc, Yca, and Ycb are also mutual admittances.
These are equal to the sums of the admittances connecting to nodes indicated in their subscripts. I1 is the
sum of all the source currents at node a. The current which drives into the node has a positive sign, while
the current driving away from the node has a negative sign.
EXAMPLE 7.4
For the circuit shown in Fig. 7.12, write the node equations by the inspection method.
Fig. 7.12
Solution The general equations are
Yaa Va 1 Yab Vb 5 I1
(7.42)
Yba Va 1 Ybb Vb 5 I2
(7.43)
254
Circuits and Networks
Consider Eq. (7.42)
1 1
1
Yaa = + +
3 j 4 − j6
The self-admittance at the node a is the sum of admittances connected to the node a.
1
1 1
Ybb =
+ +
− j6 5 j5
The self-admittance at the node b is the sum of admittances connected to the node b.
 1 

Yab = −
 − j 6 
The mutual admittance between nodes a and b is the sum of admittances connected between nodes a and
b. Similarly, Yba 52 (21/j6), the mutual admittance between nodes b and a is the sum of the admittances
connected between nodes b and a.
10 ∠0°
I1 =
3
The source current at the node a
−10 ∠30°
I2 =
= source current leaving at the node b.
5
Therefore, the nodal equations are
 
1 1

 + − 1 V −  −1V = 10 ∠0°
 3 j 4 j 6  a  j 6  b
3
(7.44)
1 1
 −1
1
−10 ∠30°
− Va +  + − Vb =
5
 5 j 5 j 6 
 j 6 
(7.45)
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 2
rrr 7-2.1
Determine the value of source currents by loop analysis for the circuit shown in Fig. Q.1 and
verify the results by using node analysis.
Fig. Q.1
rrr 7-2.2
Determine the power out of the source in the
circuit shown in Fig. Q.2 by nodal analysis and
verify the results by using loop analysis.
Fig. Q.2
255
Steady-State AC Analysis
rrr7-2.3
For the circuit shown in Fig. Q.3, obtain the voltage across 500 kV resistor.
Fig. Q.3
Frequently Asked Questions linked to LO 2
rrr 7-2.1
In the network system shown in Fig. Q.1, find the current through Z3 using nodal method. The
values of voltages are given volts and the impedances in ohms.
[BPUT 2008]
rrr 7-2.2 Using nodal analysis find voltage V1 and V2 for the circuit shown in Fig. Q.2.
[GTU Dec. 2010]
Z2 = 6–j8
2W
V2 =100––60º
V1 = 100<0º
Z1 = 6+j8
Z3 = j10
2A
4W
Fig. Q.1
rrr 7-2.3
4W
Fig. Q.2
Solve for the nodal voltage V1, V2, V3, and V4 as shown in the network in Fig. Q.3, using the nodal
analysis.
2 V dc
2 V
1
V
V
V
[GTU May 2011]
1
rrr 7-2.4
1A
2W
Find the current through the 2 V source in Fig. Q.4, using
node voltage analysis.
[GTU Dec. 2012]
2
A
3
2
2
2 A dc
1 A dc
rrr 7-2.5
Find the source voltage Vs by using nodal technique,
assume I = 5E45º A.
[JNTU Nov. 2012]
rrr 7-2.6
By applying nodal analysis for the circuit shown Fig. Q.6, determine the power output of the source
and the power in each resistor of the circuit.
[AU May/June 2013]
1W
1
1
W
4
2W
GND
Fig. Q.3
3W
j4W
–j 2 W
I
Vs
1W
3W
–j 4 W
j5W
2V
1
W
2
x
5V
Fig. Q.4
4W
Fig. Q.5
2
8W
A -j5 W
1W
-j4 W
20 30° V
Fig. Q.6
2W
256
7.5
Circuits and Networks
SUPERPOSITION THEOREM
The superposition theorem also can be used to analyse ac circuits containing more
LO 3
than one source. The superposition theorem states that the response in any element
in a circuit is the vector sum of the responses that can be expected to flow if each
source acts independently of other sources. As each source is considered, all of the
other sources are replaced by their internal impedances, which are mostly short
circuits in the case of a voltage source, and open circuits in the case of a current source. This theorem is valid
only for linear systems. In a network containing complex
impedance, all quantities must be treated as complex
numbers.
Consider a circuit which contains two sources as
shown in Fig. 7.13.
Now let us find the current I passing through
the impedance Z2 in the circuit. According to the
superposition theorem, the current due to voltage
source V ∠0° V is I1 with current source Ia ∠0° A openFig. 7.13
circuited.
I1 =
V ∠0°
Z1 + Z 2
The current due to Ia ∠0° A is I2 with voltage source V ∠0° short circuited.
∴
I 2 = I a ∠0°×
Z1
Z1 + Z2
The total current passing through the impedance Z2 is
I 5 I1 1 I2
Fig. 7.14
Fig. 7.15
EXAMPLE 7.5
Determine the voltage across the (2 1 j5) V
impedance as shown in Fig. 7.16 by using the
superposition theorem.
Fig. 7.16
Steady-State AC Analysis
257
Solution According to the superposition theorem, the current due to the 50 ∠0° V voltage source is I1 as
shown in Fig. 7.17 with current source 20 ∠30° A open-circuited.
Current I1 =
=
50 ∠0°
50 ∠0°
=
2 + j 4 + j 5 ( 2 + j 9)
50 ∠0°
= 5.42 ∠ − 77.47° A
9.22 ∠77.47
Voltage across (2 1 j5) V due to the current I1 is
Fig. 7.17
V1 5 5.42 ∠ 2 77.47° (2 1 j5)
5 (5.38) (5.42) ∠277.47° 1 68.19°
5 29.16 ∠29.28°
The current due to the 20 ∠30° A current source is I2 as shown
in Fig. 7.18, with the voltage source 50 ∠0° V short-circuited.
( j 4) V
Current I 2 = 20 ∠30°×
2
( + j 9) V
=
Fig. 7.18
∴
20 ∠30°× 4 ∠90°
9.22 ∠77.47°
I2 5 8.68 ∠120° 2 77.47° 5 8.68 ∠42.53°
Voltage across (2 1 j5) V due to the current I2 is
V2 5 8.68 ∠42.53° (2 1 j5)
5 (8.68) (5.38) ∠42.53° 1 68.19°
5 46.69 ∠110.72°
Voltage across (2 1 j5) V due to both sources is
V 5 V1 1 V2
5 29.16 ∠29.28° 1 46.69 ∠110.72°
5 28.78 2 j4.7 2 16.52 1 j43.67
5 (12.26 1 j38.97) V
Voltage across (2 1 j5) V is V 5 40.85 ∠72.53°.
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 3
rrr 7.3.1
Apply the superposition theorem to the circuit shown in Fig. Q.1 and find the current I.
Fig. Q.1
258
Circuits and Networks
Frequently Asked Questions linked to LO 3
rrr 7-3.1
Use the superposition theorem to find the current through the 4
Fig. Q.1.
resistor in the circuit shown in
[AU May/June 2013]
Fig. Q.1
rrr 7-3.2
Find the current through the capacitor of –j 5
theorem.
reactance as shown in Fig. Q.2 using the superposition
[JNTU Nov. 2012]
Fig. Q.2
rrr 7-3.3
By the superposition theorem, calculate the current the (2 + j3)
in Fig. Q.3.
impedance branch of the circuit
[PTU 2011-12]
Fig. Q.3
7.6
TH�VENIN’S THEOREM
Thèvenin’s theorem gives us a method for simplifying a given circuit. The
LO 4
Thèvenin equivalent form of any complex impedance circuit consists of an
equivalent voltage source VTh,
and an equivalent impedance Z Th,
arranged as shown in Fig. 7.19. The
values of equivalent voltage and impedance depend on the values
in the original circuit.
Though the Thèvenin equivalent circuit is not the same as its
original circuit, the output voltage and output current are the same
in both cases. Here, the Thèvenin voltage is equal to the openFig. 7.19
259
Steady-State AC Analysis
circuit voltage across the output terminals, and impedance is equal
to the impedance seen into the network across the output terminals.
Consider the circuit shown in Fig. 7.20.
Thèvenin equivalent for the circuit shown in Fig. 7.20 between
points A and B is found as follows.
The voltage across points A and B is the Thèvenin equivalent
voltage. In the circuit shown in Fig. 7.20, the voltage across A and
B is the same as the voltage across Z2 because there is no current
through Z3.
 Z

2

VTh = V 
 Z1 + Z 2 
Fig. 7.20
The impedance between points A and B with the source replaced
by short circuit is the Thèvenin equivalent impedance. In Fig.
7.20, the impedance from A to B is Z3 in series with the parallel
combination of Z1 and Z2.
Z1Z 2
Z1 + Z 2
The Thèvenin equivalent circuit is shown in Fig. 7.21.
ZTh = Z3 +
Fig. 7.21
EXAMPLE 7.6
For the circuit shown in Fig. 7.22, determine Thèvenin’s equivalent
between the output terminals.
Solution The Thèvenin voltage, VTh, is equal to the voltage across
the (4 1 j6) V impedance. The voltage across (4 1 j6) V is
V = 50 ∠0°×
= 50 ∠0°×
( 4 + j 6)
( 4 + j 6 ) + (3 − j 4 )
4 + j6
7 + j2
Fig. 7.22
7.21∠56.3°
7.28 ∠15.95°
5 50 ∠0° 3 0.99 ∠40.35°
5 49.5 ∠40.35° V
The impedance seen from terminals A and B is
= 50 ∠0°×
ZTh = ( j 5 − j 4) +
= j1 +
Fig. 7.23
(3 − j 4) ( 4 + j 6 )
3 − j 4 + 4 + j6
5∠53.13°× 7.21∠56.3°
7.28∠15.95°
5 j1 1 4.95 ∠212.78° 5 j1 1 4.83 2 j1.095
5 4.83 2 j0.095
∴ ZTh 5 4.83 ∠21.13° V
The Thèvenin equivalent circuit is shown in Fig. 7.23.
260
Circuits and Networks
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 4
rrr 7-4.1
Find the current in the 15 V resistor in the network shown in Fig. Q.1 by Thèvenin’s theorem.
Fig. Q.1
Use Thèvenin’s theorem to find the current through the (5 1 j4) V impedance in Fig. Q.2. Verify the
results using Norton’s theorem.
rrr 7-4.3 Find Thèvenin’s equivalent for the network shown in Fig. Q.3.
rrr 7-4.2
Fig. Q.2
rrr 7-4.4
Fig. Q.3
For the circuit shown in Fig. Q.4, obtain the Thèvenin’s
equivalent circuit at terminals ab.
Fig. Q.4
Frequently Asked Questions linked to LO 4
rrr 7-4.1
Find the current through the branch a-b of the network shown in Fig. Q.1 using Thevenin’s theorem.
[AU May/June 2013]
rrr 7-4.2 Find the current in the (1 + j1) W resistor across A, B of the network shown in Fig. Q.2 using
Thevenin’s theorem.
[PTU 2009-10]
5W
j5W
a
8W
10 Ð0ºV
1W
j1 W
1W
A
5W
1W
V1 = 10Ð0ºVolts
1W
j4W
b
Fig. Q.1
B
Fig. Q.2
j1 W
V2 = 10Ð60º volts
261
Steady-State AC Analysis
7.7
NORTON’S THEOREM
Another method of analysing a complex
LO 5
impedance circuit is given by Norton’s
theorem. The Norton equivalent form of
any complex impedance circuit consists
of an equivalent current source IN and
an equivalent impedance ZN, arranged as shown in Fig. 7.24. The
values of equivalent current and impedance depend on the values in
the original circuit.
Though Norton’s equivalent circuit is not the same as its original
circuit, the output voltage and current are the same in both cases;
Norton’s current is equal to the current passing through the shortcircuited output terminals and the value of impedance is equal to the
impedance seen into the network across the output terminals.
Consider the circuit shown in Fig. 7.25.
Norton’s equivalent for the circuit shown in Fig. 7.25 between
points A and B is found as follows. The current passing through points
A and B when it is short-circuited is the Norton’s equivalent current,
as shown in Fig. 7.26.
Norton’s current IN 5 V/Z1
The impedance between points A and B, with the source replaced
by a short circuit, is Norton’s equivalent impedance.
In Fig. 7.25, the impedance
from A to B,
Z2 is in parallel with Z1.
Fig. 7.24
Fig. 7.25
Fig. 7.26
∴ ZN =
Z1 Z 2
Z1 + Z 2
Norton’s equivalent circuit is shown in Fig. 7.27.
Fig. 7.27
EXAMPLE 7.7
For the circuit shown in Fig. 7.28, determine Norton’s
equivalent circuit between the output terminals, AB.
Solution Norton’s current IN is equal to the current passing
through the short-circuited terminals AB as shown in Fig.
7.29.
Fig. 7.28
Fig. 7.29
262
Circuits and Networks
The current through terminals AB is
25 ∠0°
25 ∠0°
IN =
=
3 + j 4 5 ∠53.13°
= 5 ∠−53.13°
The impedance seen from terminals AB is
ZN =
=
(3 + j 4) ( 4 − j 5)
(3 + j 4) + ( 4 − j 5)
5 ∠53.13°×6.4 ∠−51.34°
7.07 ∠−8.133°
= 4.53∠9.92°
Fig. 7.30
Norton’s equivalent circuit is shown in Fig. 7.30.
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 5
rrr 7-5.1
rrr 7-5.2
Determine Thèvenin’s and Norton’s equivalent circuits across terminals AB, in Fig. Q.1.
Determine Norton’s and Thèvenin’s equivalent circuits for the circuit shown in Fig. Q.2.
Fig. Q.1
Fig. Q.2
Frequently Asked Questions linked to LO 5
rrr 7-5.1
Find the current through the 10-ohm resistor in the following
circuit using Norton’s theorem. (Fig. Q.1)
[JNTU Nov. 2012]
A
2W
10 V
1W
10 W
5W
–j 2 W
7.8
5Ð30º V
B
MAXIMUM POWER TRANSFER THEOREM
In Chapter 3, the maximum power transfer theorem has been discussed for
resistive loads. The maximum power transfer theorem states that the maximum
power is delivered from a source to its load when the load resistance is equal to
the source resistance. It is for this reason that the ability to obtain impedance
matching between circuits is so important. For example, the audio output
transformer must match the high impedance of the audio power amplifier
Fig. Q.1
LO 6
263
Steady-State AC Analysis
output to the low input impedance of the speaker. Maximum power transfer is not always desirable, since the
transfer occurs at a 50 per cent efficiency. In many situations, a maximum voltage transfer is desired which
means that unmatched impedances are necessary. If maximum power transfer is required, the load resistance
should equal the given source resistance. The maximum power transfer theorem can be applied to complex
impedance circuits. If the source impedance is complex, then the maximum power transfer occurs when the
load impedance is the complex conjugate of the source impedance.
Consider the circuit shown in Fig. 7.31, consisting of a source impedance delivering power to a complex
load.
Current passing through the circuit shown
Vs
I=
( Rs + j X s ) + ( RL + j X L )
Vs
Magnitude of current I = I =
2
( Rs + RL ) + ( X s + X L )2
Fig. 7.31
Power delivered to the circuit is
P = I 2 RL =
Vs2
2
RL
( Rs + RL ) + ( X s + X L ) 2
In the above equation, if RL is fixed, the value of P is maximum when
Xs 52XL
Then the power P =
Vs2 RL
( Rs + RL ) 2
Let us assume that RL is variable. In this case, the maximum power is transferred when the load resistance
is equal to the source resistance (already discussed in Chapter 3). If RL 5 Rs and XL 52Xs, then ZL 5 Z *s.
This means that the maximum power transfer occurs when the load impedance is equal to the complex
conjugate of source impedance Zs.
EXAMPLE 7.8
For the circuit shown in Fig. 7.32, find the value of load
impedance for which the source delivers maximum power.
Calculate the value of the maximum power.
Solution In the circuit shown in Fig. 7.32, the maximum power
transfer occurs when the load impedance is complex conjugate
of the source impedance.
∴ ZL 5 Z *s 5 15 2 j20
When ZL 5 15 2j20, the current passing through circuit is
I=
Vs
50 ∠0°
50 ∠0°
=
= 1.66 ∠0°
=
Rs + RL 15 + j 20 + 15 − j 20
30
The maximum power delivered to the load is
P 5 I 2RL 5 (1.66)2 3 15 5 41.33 W
Fig. 7.32
264
Circuits and Networks
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 6
For the circuit shown in Fig. Q.1, find the value of Z that will receive the maximum power. Also
determine this power.
rrr 7-6.2 Determine the power output of the voltage source by loop analysis for the network shown in Fig.
Q.2. Also determine the power extended in the resistors.
rrr 7-6.1
Fig. Q.1
rrr7-6.3
rrr 7-6.4
Fig. Q.2
Obtain the Thèvenin’s equivalent circuit at terminals AB shown in Fig. Q.3.
In the circuit shown in Fig. Q.4, the resistance Rg is variable between 2 and 55 ohms. What value
of Rg results in maximum power transfer across terminals AB?
Fig. Q.3
Fig. Q.4
For the power transmission system, shown in Fig. Q.5, VS 5 240∠0°. Using PSpice, find the
average power absorbed by the load.
rrr 7-6.6 For the circuit shown in Fig. Q.6, determine the value of Z that will result in the maximum power
being delivered to Z. Calculate the value of maximum power using PSpice.
rrr 7-6.5
Fig. Q.5
rrr 7-6.7
Fig. Q.6
For the circuit shown in Fig. Q.7,
the load resistance RL is adjusted
until it absorbs the maximum
average power. Calculate the value
of RL and the maximum average
power. Use PSpice.
Fig. Q.7
265
Steady-State AC Analysis
Frequently Asked Questions linked to LO 6
r 7-6.1
In the circuit shown below, find the value of the load impedance ZL for maximum power transfer to
the load.
[AU Nov./Dec. 2012]
10 + j 10 W
V
5 0º
ZL
Fig. Q.1
Additional Solved Problems
PROBLEM 7.1
Determine (a) the equivalent voltage generator, and (b)
the equivalent current generator which may be used to
represent the given network in Fig. 7.33 at the terminals
AB.
Solution The impedance seen into the terminals when
the voltage source is short- circuited
Z AB = {[(2 || j8) − j2] || j6} = 2.99 −16.32°
Fig. 7.33
Z AB = (2.87 − j 0.84) Ω
Considering the node voltage V across j8 V and applying nodal analysis at the node, we have
V −10 0° V
V
+ +
=0
j8 j 4
2

1
1 
 = 5 0°
V  0.5 +
+

8 90° 4 90° 
∴ V = 8 36.87° V
The voltage across AB is
j6
j6
= 8 36.87°× = 12 36.87° V
j6 − j 2
j4
The current in the short-circuited terminals AB
8 36.87°
V
IA =
=
= 4 126.87° A
− j2
2 −90°
Therefore, the voltage generator is shown in Fig 7.34 (a) and the current generator is shown in Fig 7.34 (b).
VAB = V ⋅
Fig. 7.34
266
Circuits and Networks
PROBLEM 7.2
Determine the voltage Vab and Vbc in the network shown in Fig 7.35 by loop analysis, where source voltage
e(t) 5 100 cos (314t 1 45).
Fig. 7.35
Solution The circuit is redrawn as shown in Fig. 7.36.
Fig. 7.36
From Fig. 7.36 shown, the loop-current equations are given by
(3 + j8)i1 − ( j 4)i2 = 100 45°
− j 4i1 + ( j 2)i2 = 0
From the above equations, we have
3 + j8 − j 4  i1  100 45°

 =

 − j 4 j 2  i2   0 
  


1
The loop current i1 =

Where  =
1 =
3 + j8 − j 4
= j6
− j4
j2
100 45° − j 4
= 200 135°
0
j2
1
= 33.33 45° A


The loop current i2 = 2

3 + j8 100 45°
where 2 =
= 400 135°
− j4
0
∴ i1 =
∴ i2 =
2
= 66.67 45° A

(7.46)
(7.47)
267
Steady-State AC Analysis
The voltage across ab is Vab = i1 (3 + j 4) = 166.65 98.13° V
The voltage across bc is Vbc = (i1 − i2 ) j4 = 133.4 −45° V
PROBLEM 7.3
In the circuit shown in Fig 7.37, determine the power
in the impedance (2 1 j5) V connected between A
and B using Norton’s theorem.
Solution To find the current in the short-circuited
terminals AB, the circuit is redrawn as shown in Fig.
7.38.
Fig. 7.38
The total impedance in the circuit shown in Fig. 7.38,
ZT = [( 4 || − j 5) + (3 + j 5) ] V = 6.24 29.26°
The total current IT =
10 0°
= 1.6 −29.26° A
6.24 29.26°
The current I N = IT ×
− j5
= 1.24 −67.92° A
4 − j5
Open-circuit impedance seen into the terminals AB,
ZN = 4 +
(3 + j 5)(− j 5)
= (12.33 − j 5) V
3
The Norton’s equivalent circuit is shown in Fig. 7.39.
Fig. 7.39
Fig. 7.37
268
Circuits and Networks
∴ the current in (2 1 j5)V is
I ( 2+ j 5 ) V = I N ×
12.33 − j 5
12.33 − j 5 + 2 + j 5
= 1.16 −90.02° A
The power in the (2 1 j5) V impedance,
P2 1 j5 5 (1.16)2 3 2 5 2.69 watts
PROBLEM 7.4
For the circuit shown in Fig. 7.40, find the current in each resistor using the superposition theorem.
Fig. 7.40
Solution Consider the currents I1, I2, and I3 are flowing in the branches 10 V, (3 1 j4) V and (5 2 j5) V
respectively.
The branch currents due to the 100 0° voltage source can be determined by the circuit shown in Fig. 7.41,
where the 50 30° source is short-circuited.
Fig. 7.41
Total impedance ZT = [{(5 − j 5) || (3 + j 4)} + 10 ] = 14.28 4.64° V
I19 =
100 0°
= (6.97 − j 0.57) A
14.28 4.64°
1
I 29 = I1 ×
(5 − j 5) Ω
= ( 4.53 − j 4.15) A
5 − j5 + 3 + j 4
I 39 = I11 ×
(3 + j 4 )
= ( 2.45 + j 3.5) A
8 − j1
The branch current due to 50 30° V source can be determined by the circuit shown in Fig. 7.42.
Total impedance in the circuit shown in Fig 7.42.
269
Steady-State AC Analysis
ZT = [(10 || (3 + j 4)) + 5 − j 5]
= 8.46 −19.6° V
The current I 3′′ =
50 30°
= (3.83 + j 4.5) A
8.46 −19.6°
I 2′′ = I 3′′×
10
= (3.66 + j 2.33) A
10 + 3 + j 4
Fig. 7.42
3 + j4
= (0.16 + j 2.16) A
10 + 3 + j 4
The current in the 10 V resistor
I1′′= I 3′′×
I1 = I1′ − I1′′= 7.34 −21.84° A
The current in the (3 1 j4) V branch
I 2 = I 2′ + I 2′′ = 8.39 −12.53° A
The current in the (52j5) V branch
I 3 = I 3′′− I 3′ = 1.7 35.9° A
PROBLEM 7.5
Determine the maximum power delivered to the load in the circuit shown in Fig. 7.43.
Fig. 7.43
Solution The circuit is replaced by Thèvenin’s equivalent circuit in series with ZL as shown in Fig. 7.44
where VAB = I ( 3+ j 4 ) V ×(3 + j 4) volts
I ( 3+ j 4 ) V =
50 0°×(− j10)
= 34.67 −33.7° A
5 − j 6 + 3 + j 4 − j10
Voltage across AB is VAB = 34.67 −33.7°×5 53.13°
= 173.35 19.43° V
Impedance across terminals AB is
Z AB = {[(10 + j15) || (− j10)] + (5 − j 6)} || 3 + j 4
Z AB = 5.27 41.16° = 4 + j 3..5
270
Circuits and Networks
To get the maximum power delivered to the load impedance, the load
impedance must be equal to complex conjugate of source impedance.
Therefore, the total impedance in the circuit shown in Fig. 7.44 is 8 V.
The current in the circuit is
VAB 173.35
=
= 21.66 A
8
8
The maximum power transferred to the load is
I2 =
Fig. 7.44
P = I L2 RL = ( 21.66) 2 × 4 = 1874.9 watts
PROBLEM 7.6
For the circuit shown in Fig. 7.45, determine the power
output of the source and the power in each resistor of
the circuit.
Solution Assume that the voltage at the node A is VA.
By applying nodal analysis, we have
VA − 20 ∠30° VA
V
+
+ A =0
− j 4 2 + j5
3
1
1
1  20 ∠30°
VA  +
− =
 3 2 + j5 j 4 
3


VA [0.33 1 0.068 1 j0.078] 5 6.67 ∠30°
VA =
6.67 ∠30°
= 16.27 ∠18.91°
0.41∠11.09°
Current in the 2 V resistor
I2 =
VA
16.27 ∠18.91°
=
2 + j 5 5.38 ∠68.19°
∴ I2 5 3.02 ∠249.28°
Power dissipated in the 2 V resistor
P2 5 I22 R 5 (3.02)2 3 2 5 18.24 W
Current in the 3 V resistor
I3 =
−20 ∠30° + 16.27 ∠18.91°
3
5 2 6.67 ∠30° 1 5.42 ∠18.91°
5 25.78 – j3.34 1 5.13 1 j1.76 520.65 2 j1.58
I3 5 1.71 ∠2112°
Fig. 7.45
Steady-State AC Analysis
271
Power dissipated in the 3 V resistor
5 (1.71)2 3 3 5 8.77 W
Total power delivered by the source
5 VI cos f 5 20 3 1.71 cos 142° 5 26.95 W
PROBLEM 7.7
For the circuit shown in Fig. 7.46, determine the voltage VAB using the superposition theorem.
Fig. 7.46
Let the source 50 ∠0° V act on the circuit and set the source 4 ∠0° A equal to zero. If the current
source is zero, it becomes open-circuited. Then the voltage across AB is VAB 5 50 ∠0°.
Now set the voltage source 50 ∠0° V at zero, and it is short-circuited, or the voltage drop across AB is zero.
The total voltage is the sum of the two voltages.
∴ VT 5 50 ∠0°
Solution
PROBLEM 7.8
For the circuit shown in Fig. 7.47, determine the current in (2 1 j3) V by using the superposition theorem.
Fig. 7.47
Solution The current in (2 1 j3) V, when the voltage source 50 ∠0° is acting alone is
I1 =
50 ∠0°
50 ∠0°
=
(6 + j 3) 6.7 ∠26.56°
∴ I1 5 7.46 ∠226.56° A
Current in (2 1 j3) V, when the current source 20 ∠90° A is acting alone is
4
I 2 = 20 ∠90°×
(6 + j 3)
=
80 ∠90°
= 11.94 ∠63.44° A
6.7 ∠26.56°
272
Circuits and Networks
Total current in (2 1 j3) V due to both sources is
I 5 I1 1 I2
5 7.46 ∠226.56° 1 11.94 ∠63.44°
5 6.67 2 j3.33 1 5.34 1 j10.68
5 12.01 1 j7.35 5 14.08 ∠31.46°
Total current in (2 1 j3) V is I 5 14.08 ∠31.46°
PROBLEM 7.9
For the circuit shown in Fig. 7.48, determine the load current
by applying Thèvenin’s theorem.
Solution Let us find the Thèvenin equivalent circuit for the
circuit shown in Fig. 7.49 (a).
Fig. 7.48
Fig. 7.49
Voltage across AB is the voltage across ( j3) V
∴ VAB = 100 ∠0°×
= 100 ∠0°
( j 3)
( j 3) + ( j 4)
( j 3)
= 42.86 ∠0°
j7
Impedance seen from terminals AB
Z AB = ( j 5) +
( j 4) ( j 3)
j7
5 j5 1 j1.71 5 j6.71 V
Thèvenin’s equivalent circuit is shown in Fig. 7.49 (b).
If we connect a load to Fig. 7.49 (b), the current passing through the ( j5) V impedance is
IL =
42.86 ∠0°
42.86 ∠0°
= 3.66 ∠−90°
=
( j 6.71 + j 5) 11.71∠90°
273
Steady-State AC Analysis
PROBLEM 7.10
For the circuit shown in Fig. 7.50, determine the Thèvenin’s equivalent circuit.
Fig. 7.50
Solution Voltage across (2 j4) V is
V− j 4 =
=
5 ∠90°
(− j 4)
( 2 + j 2)
20 ∠0°
= 7.07 ∠−45°
2.83 ∠45°
Voltage across AB is VAB 52V10 1 V5 2 V–j4
5 –10 ∠0° 1 5 ∠90° – 7.07 ∠– 45°
5 j5 2 10 2 4.99 1 j4.99
5 214.99 1 j9.99
VAB 5 18 ∠146.31°
The impedance seen from terminals AB, when all voltage sources are short- circuited is
(2 + j 6) (− j 4)
Z AB = 4 +
2 + j2
6.32 ∠71.56°× 4 ∠−90°
= 4+
2.83 ∠45°
5 4 1 8.93 ∠263.44°
5 4 1 4 2 j 7.98 5 (8 2 j7.98) V
Fig. 7.51
Thèvenin’s equivalent circuit is shown in Fig. 7.51.
PROBLEM 7.11
For the circuit shown in Fig. 7.52, determine the load
current IL by using Norton’s theorem.
Solution Norton’s impedance seen from terminals
AB is
Z AB =
( j 3) (− j 2)
6
=
j1
( j 3) − ( j 2)
∴ ZAB 5 6 ∠290°
Fig. 7.52
274
Circuits and Networks
Current passing through AB, when it is shorted
IN =
10 ∠0°
5 ∠90°
+
3 ∠90° 2 ∠−90°
∴ IN 5 3.33 ∠290° 1 2.5 ∠180°
52j3.33 2 2.5
IN 5 4.16 ∠2126.8°
Norton’s equivalent circuit is shown in Fig. 7.53.
6 ∠−90°
Load current is I L = I N ×
5 + 6 ∠−90°
6 ∠−90°
= 4.16 ∠−126.8°×
5 − j6
4.16 ×6 ∠−216.8°
=
7.81∠−50.19°
5 3.19 ∠2166.61°
Fig. 7.53
PROBLEM 7.12
For the circuit shown in Fig. 7.54, determine Norton’s equivalent circuit.
Fig. 7.54
Solution The impedance seen from the terminals when the source is
reduced to zero,
ZAB 5 (5 1 j6) V
Current passing through the short circuited terminals, A and B,
is
Fig. 7.55
IN 5 30 ∠30° A
Norton’s equivalent circuit is shown in Fig. 7.55.
PROBLEM 7.13
Convert the active network shown in Fig. 7.56 by a single
voltage source in series with impedance.
Solution Using the superposition theorem, we can find
Thèvenin’s equivalent circuit. The voltage across AB, with 20
∠0° V source acting alone, is V9AB, and can be calculated from
Fig. 7.57 (a).
Fig. 7.56
275
Steady-State AC Analysis
Since no current is passing through the (3 1 j4) V impedance, the voltage
V 9AB 5 20 ∠0°
The voltage across AB, with 5 ∠0° A source acting alone, is V9AB, and can be calculated from Fig. 7.57 (b).
V AB 5 5 ∠0 (3 1 j4) 5 5 ∠0 3 5 ∠53.13 5 25 ∠53.13 V
Fig. 7.57
The voltage across AB, with the 10 ∠90° A source acting alone, is V AB, and can be calculated from Fig.
7.57 (c).
V AB 5 0
According to the superposition theorem, the voltage across AB due to all sources is
VAB 5 V9AB 1 V AB 1 V AB
VAB 5 20 ∠0° 1 25 ∠53.13° 5 20 1 15 1 j19.99
5 (35 1 j19.99) V 5 40.3 ∠29.73° V
The impedance seen from terminals AB
ZTh 5 ZAB 5 (3 1 j4) V
∴ the required Thèvenin circuit is shown in Fig. 7.57 (d).
Fig. 7.57
PROBLEM 7.14
For the circuit shown in Fig. 7.58, find the value of Z that will receive
maximum power; also determine this power.
Fig. 7.58
276
Circuits and Networks
Solution The equivalent impedance at terminals AB with the source set equal to zero is
Z AB =
=
5 ( j10) 7(− j 20 )
+
5 + j10 (7 − j 20)
50 ∠90°
140 ∠−90°
+
11.18 ∠63.43° 21.19 ∠−70.7°
5 4.47 ∠26.57° 1 6.6 ∠219.3°
5 3.99 1 j1.99 1 6.23 2 j2.18
5 10.22 2 j0.19
The Thèvenin equivalent circuit is shown in Fig. 7.59 (a).
The circuit in Fig. 7.59 (a) is redrawn as shown in Fig. 7.59 (b).
Fig. 7.59
Current
I1 =
=
Current
100 ∠0°
5 + j10
100 ∠0°
= 8.94 ∠− 63.43°A
11.18∠63.43°
I2 =
100 ∠0°
100 ∠0°
=
= 4.72 ∠70.7°
7 − j 20 21.19 ∠− 70.7°
Voltage at A, VA 5 8.94 ∠263.43° 3 j10 5 89.4 ∠26.57°
Voltage at B, VB 5 4.72 ∠70.7° 32j20 5 94.4 ∠219.3°
Voltage across terminals AB
VAB 5 VA 2 VB
5 89.4 ∠26.57° 2 94.4 ∠219.3°
5 79.96 1 j39.98 2 89.09 1 j31.2
5 2 9.13 1 j71.18
VTh 5 VAB 5 71.76 ∠97.3° V
To get maximum power, the load must be the complex conjugate of the source impedance.
∴ load Z 5 10.22 1 j0.19
277
Steady-State AC Analysis
Current passing through the load Z
I=
VTh
71.76 ∠97.3°
=
= 3.51∠97.3°
ZTh + Z
20.44
Maximum power delivered to the load is
5 (3.51)2 3 10.22 5 125.91 W
PROBLEM 7.15
For the circuit shown in Fig. 7.60, the resistance Rs is variable
from 2 V to 50 V. What value of Rs results in maximum power
transfer across the terminals AB?
Solution In the circuit shown, the resistance RL is fixed.
Here, the maximum power transfer theorem does not apply.
Maximum current flows in the circuit when Rs is minimum.
For the maximum current,
Rs 5 2
Fig. 7.60
But ZT 5 Rs 2 j5 1 RL 5 2 2 j5 1 20 5 (22 2 j5) 5 22.56 ∠212.8°
I=
Vs
50 ∠0°
=−
= 2.22 ∠12.8°
ZT
22.56 ∠−12.8°
Maximum power P 5 I 2R 5 (2.22)2 3 20 5 98.6 W
PROBLEM 7.16
Determine the voltage V which results in a zero current through the 2 1 j3 V impedance in the circuit shown
in Fig. 7.61.
Fig. 7.61
Solution
Choosing mesh currents as shown in Fig. 7.61, the three loop equations are
(5 1 j5) I1 2 j5 I2 5 30 ∠0°
2 j5 I1 1 (2 1 j8) I2 5 – 2V4
2 2V4 1 V4 1 V 5 0
V4 5 V
Since the current in (2 1 j3) V is zero,

I2 = 2 = 0

278
Circuits and Networks
where  2 =
5 + j 5 30 ∠0°
=0
− j 5 − 2V
(5 1 j5) (2 2V) 1 ( j5) 30 ∠0° 5 0
V=
30 ∠0°( j 5)
150 ∠90°
=
2 (5 + j 5)
14.14 ∠45°
V 5 10.608 ∠45° volts
PROBLEM 7.17
Find the value of the voltage V which results
in V0 5 5 ∠0° V in the circuit shown in Fig.
7.62.
Solution Assuming all currents are leaving
the nodes, the nodal equations are
 1
1
1 1
V
V1 
+ +  −V2   =
 5 − j 2 3 j5 
 j5  5 − j 2


 
1
1
1 
 = 2V5
−V1   + V2  +
 j5 
 j5 2 − j 2 
 


V −V 
where V5 = 5  1

 5 − j 2 
The second equation becomes
 −1
1
10 
1  −10 V
 + V2  +
=
V1  −
 j5 5 − j 2 
 j5 2 − j 2  5 − j 2




V0 = V2 =
2
= 5 ∠0°

1
1 1
+ +
5 − j 2 3 j5
−1
10
−
j5 5 − j 2
1
1 1
+ +
5 − j 2 3 j5
−1
10
−
j5 5 − j 2
V
5− j2
−10 V
5− j2
−1
j5
= 5 < 0°
1
1
+
j5 2 − j 2
The source voltage V 5 2.428 ∠288.74° volts.
Fig. 7.62
279
Steady-State AC Analysis
PROBLEM 7.18
For the circuit shown in Fig. 7.63, find the current in the
j 2 V inductance by using Thèvenin’s theorem.
Solution From the circuit shown in Fig. 7.63, the open-
circuit voltage at terminals a and b is
Voc 529 Vi
where Vi 529Vi 2 100 ∠0°
Fig. 7.63
10Vi 52100 ∠0°
Vi 5210 ∠0°
Thèvenin’s voltage Voc 5 90 ∠0°
From the circuit, short-circuit current is determined by shorting terminals a and b. Applying Kirchhoff’s
voltage law, we have
9Vi 2 j10 isc 5 0
isc 5 9∠90°
Voc 90 ∠0°
=
= 10 ∠− 90°
I sc
9 ∠90°
ZTh 52 j 10 V
The Thèvenin’s equivalent circuit is shown in Fig. 7.64.
90∠0°
The current in the j 2 V inductor is
j8
5 11.25 ∠90°
∴ ZTh =
Fig. 7.64
PROBLEM 7.19
For the circuit shown in Fig. 7.65, find the value of Z that
will receive maximum power; also determine this power.
Solution The equivalent impedance can be obtained by
finding Voc and isc at terminals a b. Assume that current i
is passing in the circuit.
i=
=
100 ∠0° − 5V4
4 + j10
100 ∠0°
5× 4i
−
4 + j10 4 + j10
i 5 3.85 ∠222.62°
Voc 5 100 ∠0° 2 4 3 3.85 ∠222.62°
5 86 ∠3.94°
isc 5 25 1 j50 5 56 ∠63.44°
Thèvenin’s equivalent impedance
V
ZTh = oc = 1.54 ∠− 59.5°
isc
5 0.78 – j1.33
Fig. 7.65
280
Circuits and Networks
The circuit is drawn as shown in Fig. 7.66.
To get maximum power, the load must be the complex conjugate
of the source impedance.
∴ load Z 5 0.78 1 j 1.33
Current passing through the load Z
I=
VTh 86 ∠3.94°
= 55.13 ∠3.94°
ZTh Z 1.56
Maximum power delivered to the load is
(55.13)2 3 (0.78) 5 2370.7 W.
Fig. 7.66
PSpice Problems
PROBLEM 7.1
For the circuit shown in Fig. 7.67, determine the power
output of the source and the power in each resistor.
* AC ANALYSIS
VS 1
0
AC 20 30
R1 1
2
3
R2 2
3
2
C1 2
0
795.77 U
L1 3
0
15.9 M
.AC LIN 1 50 100
.PRINT AC IM(VS) IP(VS) IM(R1) IP(R1)
1 VM(R1) VP(R1) IM(R2) IP(R2) VM(R2)
1 VP(R2)
.END
Fig. 7.67
OUTPUT
**** AC ANALYSIS TEMPERATURE 5 27.000 DEG C
*******************************************************************
FREQ
IM(VS)
IP(VS)
IM(R1)
IP(R1)
VM(R1)
5.000E 1 01 1.688E 1 00 2 1.126E 1 02
FREQ
VP(R1)
5.000E 1 01 6.738E 1 01
IM(R2)
1.688E 1 00
IP(R2)
6.738E 1 01 5.065E 1 00
VM(R2)
VP(R2)
3.023E 1 00 – 4.908E 1 01 6.047E 1 00 2 4.908E 1 01
Result Power dissipated in R1 5 Re{5.065∠67.38 3 1.688∠267.35}
5 Re{V(R1) 3 I(R1)*} 5 8.55 W
Power dissipated in R2 5 Re{6.049∠– 49.08 3 3.023∠49.08} 5 Re{V(R2) 3 I(R2)*}.
5 18.28 W
Power output of source 5 Re{20∠30° 3 1.688∠112.6°} 5 26.82 W 5 Re {Vs 3 Is*}.
281
Steady-State AC Analysis
PROBLEM 7.2
Using PSpice, for the circuit shown in Fig. 7.68, determine Thèvenin’s equivalent circuit.
Fig. 7.68
* NETLIST TO FIND VTH
V1 1
3
AC 5 90
R1 1
2
2
L1 2
0
19.108 MH
C1 3
0
796.18 UF
R2 1
4
4
V2 4
5
AC 10 0
.AC LIN 1 50 50
.PRINT AC VM(5,0) VP(5,0) IM(V2) IP(V2)
.END
* NETLIST TO FIND ZTH
V1
1
3 AC 0 0
R1
1
2 2
L1
2
0 19.108 MH
C1
3
0 796.18 UF
R2
1
4 4
V2
4
5 AC 0 0
VX
0
5 AC 1 0
.AC LIN 1 50 50
.PRINT AC IM(VX) IP(VX)
Fig. 7.69
.END
**** AC ANALYSIS TEMPERATURE 5 27.000 DEG C
****************************************************************
FREQ
VM(5,0)
VP(5,0)
IM(V2)
IP(V2)
5.000E 1 01
1.801E 1 01
1.463E 1 02
1.000E 2 30
0.000E 1 00
**** AC ANALYSIS TEMPERATURE 5 27.000 DEG C
****************************************************************
FREQ
IM(VX)
IP(VX)
5.000E 1 01
8.850E 2 02
21.350E 1 02
Result
Vth 5 V(5) 5 180.1∠146.3°
Zth 5 2 1/I(VX) 5 1/0.0850∠135° 5 8.318∠– 45°
282
Circuits and Networks
Answers to Practice Problems
7-1.1
3.39 ∠297.3°
7-1.2
I 2+ j 3 = 1.74 40.1° A
7-1.4
I 5 0.25 A
7-3.1
I 5 OA
7-4.1
4.37 A
7-4.3
(20.18 2 j0.6)V1 volts in series with (100 2 j30) V
7-4.4
0.894 ∠263.4° in series with (0.4 1 j1.25) V
(1.1 1 j4.7) V in series with (0.93 1 j0.75) V
7-5.1
(3.2 1 j2.4) A in parallel with (0.93 1 j0.75) V
7-6.1
(3.82 2 j1.03) V; 15.11 W
7-6.3 The voltage source 11.39 264.4° V is in series with impedance (10.97 − j2.16) V
7-6.4
PAB 5 593 watts
Objective-Type Questions
rrr 7.1
The superposition theorem is valid
(a) only for ac circuits
(c) For both, ac and dc circuits
(b) only for dc circuits
(d) neither of the two
rrr 7.2
When applying the superposition theorem to any circuit,
(a) the voltage source is shorted, the current source is opened
(b) the voltage source is opened, the current source is shorted
(c) both are opened
(d) both are shorted
rrr 7.3
While applying Thèvenin’s theorem, the Thèvenin’s voltage is equal to
(a) short-circuit voltage at the terminals
(b) open-circuit voltage at the terminals
(c) voltage of the source
(d) total voltage available in the circuit
rrr 7.4 Thèvenin impedance ZTh is found
(a) by short-circuiting the given two terminals
(b) between any two open terminals
(c) by removing voltage sources along with the internal
resistances
(d) between same open terminals as for VTh
rrr 7.5
Thèvenin impedance of the circuit at its terminals A and
B in Fig. 7.70 is
(a) 5 H
(c) 1.4 V
rrr 7.6
(b) 2 V
(d) 7 H
Norton’s equivalent form in any complex impedance circuit consists of
(a) an equivalent current source in parallel with an equivalent resistance
(b) an equivalent voltage source in series with an equivalent conductance
Fig. 7.70
283
Steady-State AC Analysis
(c) an equivalent current source in parallel with an equivalent impedance
(d) None of the above
rrr 7.7
The maximum power transfer theorem can be applied
(a) only to dc circuits
(c) to both dc and ac circuits
(b) only to ac circuits
(d) neither of the two
rrr 7.8
In a complex impedance circuit, the maximum power transfer occurs when the load impedance is equal to
(a) complex conjugate of source impedance
(b) source impedance
(c) source resistance
(d) none of the above
rrr 7.9
Maximum power transfer occurs at a
(a) 100% efficiency
(c) 25% efficiency
(b) 50% efficiency
(d) 75% efficiency
rrr 7.10 In the circuit shown in Fig. 7.71, the power supplied by
the 10 V source is
(a) 6.6 W
(c) 30 W
(b) 21.7 W
(d) 36.7 W
Fig. 7.71
rrr 7.11 The Thèvenin equivalent impedance of the circuit in
Fig. 7.72 is
(a) (1 1 j5) V
(c) (6.25 1 j6.25) V
(b) (2.5 1 j25) V
(d) (2.5 1 j6.25) V
rrr 7.12 A source has an emf of 10 V and an impedance of
500 1 j100 V. The amount of maximum power
transferred to the load will be
(a) 0.5 mW
(c) 0.05 W
Fig. 7.72
(b) 0.05 mW
(d) 0.5 W
rrr 7.13 For the circuit shown in Fig. 7.73, find the voltage
across the dependent source.
(a) 8 ∠0°
(c) 4 ∠90°
(b) 4 ∠0°
(d) 8 ∠290°
Fig. 7.73
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8
LEARNING OBJECTIVES
8.1
SERIES RESONANCE
In many electrical circuits, resonance is a very important phenomenon.
LO 1
The study of resonance is very useful, particularly in the area of
communications. For example, the ability of a radio receiver to select a
certain frequency, transmitted by a station and to eliminate frequencies
from other stations is based on the principle of resonance. In a series RLC
circuit, the current lags behind, or leads the applied voltage depending
upon the values of XL and XC. XL causes the total current to lag behind the applied voltage, while
XC causes the total current to lead the applied voltage. When XL > XC, the circuit is predominantly
inductive, and when XC > XL, the circuit is predominantly capacitive. However, if one of the parameters
of the series RLC circuits is varied in such a way that the current in the circuit is in phase with the
applied voltage, then the circuit is said to be in resonance.
Consider the series RLC circuit shown in Fig. 8.1.
The total impedance for the series RLC circuit is

1 
Z = R + j ( X L − X C ) = R + j vL −


vC 
Fig. 8.1
It is clear from the circuit that the current I 5 VS /Z.
The circuit is said to be in resonance if the current is in phase
285
Resonance
with the applied voltage. In a series RLC circuit, series resonance occurs when XL 5 XC. The frequency at which
the resonance occurs is called the resonant frequency.
Since XL 5 XC, the impedance in a series RLC circuit is purely resistive. At the resonant frequency, fr, the
voltages across capacitance and inductance are equal in magnitude. Since they are 180° out of phase with
each other, they cancel each other and, hence zero voltage appears across the LC combination.
At resonance,
1
XL = X C , i.e. , vL =
vC
Solving for resonant frequency, we get
1
2pf r L =
2p f r C
1
f r2 =
2
4p LC
1
∴ fr =
2p LC
In a series RLC circuit, resonance may be produced by varying the frequency, keeping L and C constant;
otherwise, resonance may be produced by varying either L or C for a fixed frequency.
EXAMPLE 8.1
For the circuit shown in Fig. 8.2, determine the value of capacitive
reactance and impedance at resonance.
Solution At resonance,
XL 5 XC
Since XL 5 25 V
1
= 25
vC
The value of impedance at resonance is
X C = 25 V
∴
Fig. 8.2
Z5R
Z 5 50 V
EXAMPLE 8.2
Determine the resonant frequency for the circuit shown in Fig. 8.3.
Solution The resonant frequency is
fr =
1
2p LC
1
2p 10 ×10−6 × 0.5×10−3
fr 5 2.25 kHz
Fig. 8.3
286
Circuits and Networks
Frequently Asked Questions linked to LO 1*
rrr 8-1.1
rrr 8-1.2
8.2
[AU April/May 2011]
[RGTU June 2014]
When do you say that a given ac circuit is at resonance?
What do you understand by resonance?
IMPEDANCE AND PHASE ANGLE OF A SERIES RESONANT CIRCUIT
The impedance of a series RLC circuit is
LO 2
2

1 
Z = R 2 + vL −


vC 
The variation of XC and XL with frequency is shown in Fig. 8.4.
Fig. 8.4
At zero frequency, both XC and Z are infinitely large, and XL is zero because at zero frequency, the capacitor
acts as an open circuit and the inductor acts as a short circuit. As the frequency increases, XC decreases and XL
increases. Since XC is larger than XL, at frequencies below the resonant frequency fr, Z decreases along with
XC. At resonant frequency fr, XC 5 XL, and Z 5 R. At frequencies above the resonant frequency fr, XL is larger
than XC, causing Z to increase. The phase angle as a function of frequency is shown in Fig. 8.5.
At a frequency below the resonant frequency, current leads the source voltage because the capacitive
reactance is greater than the inductive reactance. The phase angle decreases as the frequency approaches the
resonant value, and is 0° at resonance. At frequencies above resonance, the current lags behind the source
voltage, because the inductive reactance is greater than capacitive reactance. As the frequency goes higher,
the phase angle approaches 90°.
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
287
Resonance
Fig. 8.5
EXAMPLE 8.3
For the circuit shown in Fig. 8.6, determine the impedance at
resonant frequency, 10 Hz above resonant frequency, and 10 Hz
below resonant frequency.
Solution Resonant frequency
fr =
=
1
2p LC
Fig. 8.6
1
2p 0.1×10 ×10−6
= 159.2 Hz
At 10 Hz below fr 5 159.2 2 10 5 149.2 Hz
At 10 Hz above fr 5 159.2 1 10 5 169.2 Hz
Impedance at resonance is equal to R
∴ Z 5 10 V
Capacitive reactance at 149.2 Hz is
1
1
X C1 =
=
v1C 2p×149.2×10−6 ×10
∴ XC1 5 106.6 V
Capacitive reactance at 169.2 Hz is
1
1
X C2 =
=
v2C 2p×169.2×10 ×10−6
∴ XC2 5 94.06 V
288
Circuits and Networks
Inductive reactance at 149.2 Hz is
XL1 5 v2L 5 2p 3 149.2 3 0.1 5 93.75 V
Inductive reactance at 169.2 Hz is
XL2 5 v2L 5 2p 3 169.2 3 0.1 5 106.31 V
Impedance at 149.2 Hz is
Z = R 2 + ( X L1 − X C1 ) 2
= (10) 2 + (93.75 −106.6) 2
= 16.28 V
Here XC1 is greater than XL1, so Z is capacitive.
Impedance at 169.2 Hz is
Z = R 2 + ( X L2 − X C2 ) 2
= (10) 2 + (106.31− 94.06) 2
= 15.81 V
Here, XL1 is greater than XC1, so Z is inductive.
Frequently Asked Questions linked to LO 2
rrr 8-2.1
rrr 8-2.2
8.3
For the circuit shown in Fig. Q.1, determine the impedance
at resonant frequency, 10 Hz above resonant frequency,
and 10 Hz below resonant frequency.
[AU May/June 2014]
Define the following:
[JNTU Nov. 2012]
(a) Impedance (b) Phase angle
10 W
0.1 H
Vs
Fig. Q.1
VOLTAGES AND CURRENTS IN A SERIES RESONANT CIRCUIT
The
variation
of
impedance
and current with
frequency
is
shown in Fig.
8.7.
At resonant
frequency, the capacitive reactance is equal to
inductive reactance, and hence the impedance
is minimum. Because of minimum impedance,
maximum current flows through the circuit. The
current variation with frequency is plotted.
LO 3
Fig. 8.7
10nF
289
Resonance
The voltage drop across resistance, inductance and capacitance also varies with frequency. At f 5 0, the
capacitor acts as an open circuit and blocks current. The complete source voltage appears across the capacitor.
As the frequency increases, XC decreases and XL increases, causing total reactance XC – XL to decrease. As
a result, the impedance decreases and the current increases. As the current increases, VR also increases, and
both VC and VL increase.
When the frequency reaches its resonant value fr, the impedance is equal to R, and hence, the current
reaches its maximum value, and VR is at its maximum value.
As the frequency is increased above resonance, XL continues to increase and XC continues to decrease,
causing the total reactance, XL – XC to increase.
As a result there is an increase in impedance and
a decrease in current. As the current decreases,
VR also decreases, and both VC and VL decrease.
As the frequency becomes very high, the current
approaches zero, both VR and VC approach zero,
and VL approaches Vs.
The response of different voltages with
frequency is shown in Fig. 8.8.
The drop across the resistance reaches its
maximum when f 5 fr. The maximum voltage
across the capacitor occurs at f 5 fc. Similarly, the
maximum voltage across the inductor occurs at
f 5 fL.
The voltage drop across the inductor is
VL 5 IXL
Fig. 8.8
V
where I =
Z
∴ VL =
vLV
2

1 
R 2 + vL −


vC 
To obtain the condition for maximum voltage across the inductor, we have to take the derivative of the above
equation with respect to frequency, and make it equal to zero.
dVL
∴
=0
dv
If we solve for v, we obtain the value of v when VL is maximum.



2 −1/ 2 



dVL
d 
1  
2



=

vLV  R + vL −



dv
dv 
vC  







−1/ 2

 2
2
1
L

LV  R + v2 L2 −
+

C v2C 2 
1 
2 
vLV  2
2L
2 2
−
+ 2 2 2vL2 − 3 2 
 R + v L −

C v C 
2 
v C =0
2
1
L
+
R 2 + v2 L2 −
C v 2C 2
290
Circuits and Networks
From this,
2L
+ 2 / v 2C 2 = 0
C
2
1
∴ vL =
=
2 LC − R 2C 2
LC
R2 −
fL =
1
2
2−
R 2C
L
1
2p LC
1−
R 2C
2L
Similarly, the voltage across the capacitor is
VC = IX C =
∴ VC =
1
vC
V
2

1 
R 2 + vL −


vC 
To get maximum value
×
1
vC
dVC
=0
dv
If we solve for v, we obtain the value of v when VC is maximum.
2 −1/ 2
 

dVC
1 
1
1  
1 
 2 vL −
L + 2 
= vC  R 2 + vL −







 
2
vC  
vC 
dv
v C 



2

1 
+ R 2 + vL −
C =0


vC 
From this,
vC2 =
vC =
∴
fC =
1
R2
−
LC 2 L
1
R2
−
LC 2 L
1
1
R2
−
2p LC 2 L
The maximum voltage across the capacitor occurs below the resonant frequency; and the maximum
voltage across the inductor occurs above the resonant frequency.
EXAMPLE 8.4
A series circuit with R 5 10 V, L 5 0.1 H and C 5 50 mF has an applied voltage V 5 50∠0° with a
variable frequency. Find the resonant frequency, the value of frequency at which maximum voltage occurs
across the inductor and the value of frequency at which maximum voltage occurs across the capacitor.
291
Resonance
Solution The frequency at which maximum voltage occurs across the inductor is
fL =
=
1
1
 R 2C 
1−


2 L 

2p LC
1
−6
2p 0.1×50 ×10
1
 10 2 ×50 ×10−6 
( )

1− 


2
×
0
.
1


5 72.08 Hz
Similarly,
fC =
R2
1
1
−
2p LC 2 L
2
(10)
1
1
=
−
6
−
2× 0.1
2p 0.1×50 ×10
1
200000 − 500
2p
5 71.08 Hz
1
1
=
= 71.18 Hz
Resonant frequency f r =
2p LC
2p 0.1×50 ×10−6
It is clear that the maximum voltage across the capacitor occurs below the resonant frequency and the
maximum inductor voltage occurs above the resonant frequency.
=
8.4
BANDWIDTH OF AN RLC CIRCUIT
LO 3
The bandwidth of any system is the range of frequencies for which the current or output voltage is equal to
70.7% of its value at the resonant frequency, and it is denoted by BW. Figure 8.9 shows the response of a
series RLC circuit.
The frequency f1 is the frequency at
which the current is 0.707 times the current
at resonant value, and it is called the
lower cut-off frequency. The frequency f2
is the frequency at which the current is
0.707 times the current at resonant value
(i.e. maximum value), and is called the
upper cut-off frequency. The bandwidth,
or BW, is defined as the frequency
difference between f2 and f1.
∴ BW 5 f2 – f1
The unit of BW is hertz (Hz).
Fig. 8.9
292
Circuits and Networks
If the current at P1 is 0.707Imax, the impedance of the circuit at this point is 2R, and hence
1
− v1 L = R
v1C
Similarly,
v2 L −
1
=R
v 2C
(8.1)
(8.2)
If we equate both the above equations, we get
1
1
− v1 L = v2 L −
v1C
v 2C
L(v1 + v2 ) =
1  v1 + v2 


C  v1v2 
(8.3)
From Eq. (8.3), we get
v1v2 =
1
LC
1
LC
∴ v2r 5 v1v2
we have v2r =
(8.4)
If we add Eqs (8.1) and (8.2), we get
1
1
= 2R
− v1 L + v2 L −
v1C
v 2C
1  v2 − v1 
 = 2 R

C  v1v2 
1
Since C = 2
vr L
(v2 − v1 ) L +
and
(v2 − v1 ) L +
(8.5)
v1v2 5 v2r
v2r L(v2 − v1 )
v2r
= 2R
(8.6)
From Eq. (8.6), we have
R
L
(8.7)
f 2 − f1 =
R
2pL
(8.8)
BW =
R
2pL
v2 − v1 =
∴
or
Resonance
293
From Eq. (8.8), we have
R
f 2 − f1 =
2pL
R
∴ f r − f1 =
4 pL
R
f2 − fr =
4 pL
The lower frequency limit
The upper frequency limit
R
4 pL
R
f2 = fr +
4 pL
f1 = f r −
(8.9)
(8.10)
If we divide the equation on both sides by fr, we get
f 2 − f1
R
=
(8.11)
fr
2pf r L
Here, an important property of a coil is defined. It is the ratio of the reactance of the coil to its resistance.
This ratio is defined as the Q of the coil. Q is known as a figure of merit, it is also called quality factor and is an
indication of the quality of a coil.
Q=
X L 2pf r L
=
R
R
(8.12)
If we substitute Eq. (8.11) in Eq. (8.12), we get
f 2 − f1
1
(8.13)
=
fr
Q
The upper and lower cut-off frequencies are sometimes called the half-power frequencies. At these
frequencies, the power from the source is half of the power delivered at the resonant frequency.
At resonant frequency, the power is
Pmax 5 I 2max R
I
2
I2 R
At frequency f1 , the power is P1 =  max  R = max
 2 
2
Similarly, at frequency f2, the power is
I
2
P2 =  max  R
 2 
I2 R
= max
2
The response curve in Fig. 8.9 is also called the selectivity curve of the circuit. Selectivity indicates how
well a resonant circuit responds to a certain frequency and eliminates all other frequencies. The narrower the
bandwidth, the greater the selectivity.
EXAMPLE 8.5
Determine the quality factor of a coil for the series circuit consisting of R 5 10 V, L 5 0.1 H and C 5 10 mF.
294
Circuits and Networks
Solution
Quality factor Q =
fr =
fr
BW
1
2p LC
=
1
2p 0.1×10 ×10−6
= 159.2 Hz
At lower half-power frequency, XC > XL
1
− 2pf1 L = R
2pf1C
f1 =
From which
−R + R 2 + 4 L / C
4 pL
At upper half-power frequency, XL > XC
1
2pf 2 L −
=R
2p f 2 C
From which
Bandwidth
+R + R2 + 4 L / C
4 pL
R
BW = f 2 − f1 =
2pL
f2 =
fr
2pf r L 2× p×159.2× 0.1
=
=
BW
R
10
fr
Q0 =
= 10
BW
Hence Q0 =
8.5
THE QUALITY FACTOR (Q) AND ITS EFFECT ON BANDWIDTH
LO 3
The quality factor, Q, is the ratio of the reactive power in the inductor or capacitor to the true power in the
resistance in series with the coil or capacitor.
The quality factor
maximum energy stored
Q = 2p×
energy dissipated per cycle
2
In an inductor, the maximum energy stored is given by LI
2
2
2
 I 
I RT
Energy dissipated per cycle =   R ×T =
 2 
2
1 2
LI
∴ Quality factor of the coil Q = 2p× 22
I R 1
×
2
f
2p f L v L
=
=
R
2
2
Similarly, in a capacitor, the maximum energy stored is given by CV
2
The energy dissipated per cycle 5 ( I / 2 ) 2 R × T
295
Resonance
The quality factor of the capacitance circuit
Q=
Q=
In series circuits, the quality factor
2
1  1 
2p C 

2  vC 
2
I
1
R×
2
f
=
1
vCR
vL
1
=
R
vCR
fr
.
BW
A higher value of the circuit Q results in a smaller bandwidth. A lower value of Q causes a larger bandwidth.
We have already discussed the relation between bandwidth and quality factor, which is Q =
EXAMPLE 8.6
For the circuit shown in Fig. 8.10, determine the value
of Q at resonance and bandwidth of the circuit.
Solution The resonant frequency,
fr =
=
1
2p LC
Fig. 8.10
1
2p 5×100 ×10−6
5 7.12 Hz
Quality factor Q 5 XL/R 5 2pfr L/R
=
2p× 7.12×5
= 2.24
100
Bandwidth of the circuit is BW =
8.6
fr
7.12
=
= 3.178 Hz
Q 2.24
MAGNIFICATION IN RESONANCE
LO 3
If we assume that the voltage applied to the series RLC circuit is V, and the current at resonance is I, then the
voltage across L is VL 5 IXL 5 (V/R) vrL
Similarly, the voltage across C
VC = IX C =
V
Rvr C
Since Q 5 1/vrCR 5 vr L/R
where vr is the frequency at resonance.
296
Circuits and Networks
Therefore, VL 5 VQ
VC 5 VQ
The ratio of voltage across either L or C to the voltage applied at resonance can be defined as magnification.
magnification 5 Q 5 VL / V or VC / V
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 3
rrr 8-3.1
rrr 8-3.2
rrr 8-3.3
rrr 8-3.4
rrr 8-3.5
A voltage v(t) 5 50 sin vt is applied to a series RLC circuit. At the resonant frequency of the
circuit, the maximum voltage across the capacitor is found to be 400 V. The bandwidth is known
to be 500 rad/sec and the impedance at resonance is 100 V. Find the resonant frequency, and
compute the upper and lower limits of the bandwidth.
Determine the values of L and C of the circuit.
For the circuit shown in Fig. Q.2, determine the
frequency at which the circuit resonates. Also find
the voltage across the capacitor at resonance, and the
Q-factor of the circuit.
A series RLC circuit has a quality factor of 10 at
200 rad/s. The current flowing through the circuit at
resonance is 0.5 A and the supply voltage is 10 V. The
Fig. Q.2
total impedance of the circuit is 40 V. Find the circuit
constants.
An RLC series circuit is to be chosen to produce a
magnification of 10 at 100 rad/s. The source can
supply a maximum current of 10 A and the supply
voltage is 100 V. The power frequency impedance of
the circuit should not be more than 14.14 V. Find the
values of R, L, and C.
For the circuit shown in Fig. Q.5, the applied voltage
v(t) 515 sin 1800 t. Determine the resonant frequency.
Fig. Q.5
Calculate the quality factor and bandwidth. Compute
the lower and upper limits of the bandwidth.
rrr 8-3.6
In the circuit shown in Fig. Q.6, the
current is at its maximum value with
inductor value L 5 0.5 H, and 0.707
times of its maximum value with
L 5 0.2 H. Find the value of Q at
v 5 200 rad/ s and circuit constants.
rrr 8-3.7
In a series RLC circuit, if the applied
Fig. Q.6
voltage is 10 V, what is the maximum
voltage across the inductor, if the resonance frequency is 1 kHz, and Q-factor is 10.
rrr 8-3.8
Obtain the expression for the frequency at which the maximum voltage occurs across the capacitor
in series resonance circuit in terms of Q-factor and resonance frequency.
297
Resonance
The voltage applied to the series RLC circuit is 5 V. The Q of the coil is 25 and the value of
the capacitor is 200 PF. The resonant frequency of the circuit is 200 kHz. Find the value of
inductance, the circuit current and the voltage across the capacitor.
rr 8-3.10 The resonance frequency of a series RLC circuit is 1 kH; the quality factor is 10, the source
voltage is 10 V. (a) Find the voltage across the conductor at resonance (b) Also, find the frequency
for which the voltage across the inductor is maximum.
r 8-3.9
Frequently Asked Questions linked to LO 3
r 8-3.1
r 8-3.2
r 8-3.3
r 8-3.4
r 8-3.5
r 8-3.6
r 8-3.7
rr 8-3.8
rr 8-3.9
rrr8-3.10
r 8-3.11
r 8-3.12
r 8-3.13
r 8-3.14
r 8-3.15
r 8-3.16
rr 8-3.17
Draw the frequency response of an RLC series circuit.
Define bandwidth of a resonant circuit.
[AU May/June 2013]
An RLC series circuit consists of R = 16 W, L = 5 mH, and C = 2 mF. Calculate the quality factor
at resonance, bandwidth, and half-power frequencies.
[AU May/June 2014]
State the concept of bandwidth of a series RLC circuit.
[AU Nov./Dec. 2012]
Calculate the halfpower frequencies of a series resonant circuit where the resonance frequency is
250 × 103 Hz and bandwidth is 150 kHz.
[BPUT 2007]
Under what condition will the power in a series RLC circuit will half that at resonance?
[BPTU 2008]
Derive the expression for bandwidth of a series RLC circuit.
[JNTU Nov. 2012]
A series RLC circuit has the following parameters: R = 15 ohms, L = 2 H, C = 100 micro F.
Calculate the resonant frequency. Under resonant condition, calculate current, power, and voltage
drops across various elements if the applied voltage is 100 V.
[JNTU Nov. 2012]
In a series resonancetype bandpass filter, L = 60 mHz, C = 150 nF, and R = 70 W. [PTU 2011-12]
Determine
(a) Resonance frequency in Hz, (b) Bandwidth, (c) Cut-off frequencies
Assume the load resistance to be 600 W.
A capacitor of 400 pF is in series with a coil resonant at 1 MHz. The halfpower frequencies are
0.9 MHz. Specify values of R and L. Also find second halfpower frequency.
[PU 2010]
Derive the expression for bandwidth of a series resonant circuit.
[PU 2012]
Define quality factor in the resonant circuit.
[AU May/June 2014]
Determine the quality factor of a coil for series resonant circuit of R = 10 ohms, L = 0.1 H, and
C = 10 microfarads.
[AU May/June 2014]
Determine the quality factor of a coil for the series circuit consisting of R = 10 , L = 0.1 H, and
C = 10 F. Derive the formula used.
[AU April/May 2011]
Derive the expression for the impedance of a series resonating circuit in the terms of Q0 and .
[PU 2010]
A series RLC circuit consists of a resistance of 1 k and an inductance of 100 mH in series with
a capacitance of 10 pF. If 100 V is applied at input across the combination, determine [PU 2010]
(a) Resonant frequency
(b) Maximum current in the circuit
(c) Quality factor of the circuit
(d) Half-power frequencies.
R
L
C
V
rr
L
r
RLC
mF
298
Circuits and Networks
Q
RLC
r
R
L
C
Q
r
Q
Q0
r
8.7
PARALLEL RESONANCE
Basically, parallel resonance occurs when XC 5 XL. The frequency at
which resonance occurs is called the resonant frequency. When XC 5 XL,
the two branch currents are equal in magnitude and 180° out of phase with
each other. Therefore, the two currents cancel each other out, and the total
current is zero. Consider the circuit shown in Fig. 8.11. The
condition for resonance occurs when XL 5 XC.
In Fig. 8.11, the total admittance
1
1
Y=
+
RL + jvL RC − ( j / vC )
=
RL − jvL
RL2
2 2
+v L
+
RC + ( j / vC )
1
RC2 + 2 2
vC






 

 

L
1
/
v
v
C


 
= 2
+
+ j
−


1   RL2 + v2 L2  
 2
RL + v2 L2 R 2 + 1


 RC + 2 2  

C


v 2C 2
v C 


At resonance, the susceptance part becomes zero.
1
vr L
vr C
∴
=
1
2
RL2 + v2r L2
RC + 2 2
vr C


1
1
 R 2 + v2 L2 
vr L  RC2 + 2 2  =
r
 L

v
C
v
C
r


r

1 
1  2
2 2
vr2  RC2 + 2 2  =
 RL + vr L 
LC
vr C 

v2 L
1 2
1
RL − 2
v2r RC2 − r =
C
LC
C



1
L
L
 RL2 − 
v2r  RC2 −  =



C  LC 
C 
RL
RC
LO
4
Fig. 8.11
(8.14)
(8.15)
299
Resonance
vr =
1
RL2 − ( L / C )
LC
RC2 − ( L / C )
(8.16)
The condition for resonant frequency is given by Eq. (8.16.) As a special case, if RL 5 RC, then Eq. (8.16)
becomes
1
vr =
LC
1
Therefore f r =
2p LC
EXAMPLE 8.7
Find the resonant frequency in the ideal parallel LC
circuit shown in Fig. 8.12.
1
Solution f r =
2p LC
1
=
−3
2p 50 ×10 × 0.01×10−6
= 7117.6 Hz
8.8
RESONANT FREQUENCY FOR A TANK CIRCUIT
Fig. 8.12
LO 4
The parallel resonant circuit is generally called a tank circuit
because of the fact that the circuit stores energy in the magnetic
field of the coil and in the electric field of the capacitor. The stored
energy is transferred back and forth between the capacitor and coil
and vice-versa. The tank circuit is shown in Fig. 8.13. The circuit
is said to be in resonant condition when the susceptance part of
admittance is zero.
The total admittance is
1
1
Y=
+
(8.17)
RL + jX L − jX C
Fig. 8.13
Simplifying Eq. (8.17), we have
Y=
RL − jX L
RL2
+
X L2
+
j
XC

 1
X
+ j 
− 2 L 2 
+
 X C RL + X L 
To satisfy the condition for resonance, the susceptance part is zero.
X
1
∴
= 2 L 2
XC
RL + X L
=
RL
RL2
X L2
(8.18)
300
Circuits and Networks
vC =
vL
(8.19)
RL2 + v2 L2
From Eq. (8.19), we get
RL2 + v2 L2 =
v2 L2 =
v2 =
∴ v=
L
C
L
− RL2
C
R2
1
− 2L
LC L
R2
1
− 2L
LC L
(8.20)
The resonant frequency for the tank circuit is
fr =
R2
1
1
− 2L
2p LC L
(8.21)
EXAMPLE 8.8
For the tank circuit shown in Fig. 8.14, find the resonant frequency.
Solution The resonant frequency
fr =
R2
1
1
− 2L
2p LC L
2
=
(10)
1
1
−
−
6
2
2p 0.1×10 ×10
(0.1)
=
1
2p
6
2
(10) − (10) =
Fig. 8.14
1
(994.98) = 158.35 Hz
2p
Frequently Asked Questions linked to LO 4
fr
fr
rrr
rrr
RC
RL
V
L
C
V
Fig. Q.1
XC
RL
RL
Fig. Q.2
XL
301
Resonance
R
rr
X
rr
RL
10 W
5W
j10 W
8W
–j XC
–j 2 W
j3 W
Fig. Q.3
Fig. Q.4
r
r
fr
r
L
rr
W
RL
jXL
V
5W
–jXC
Fig. Q.7
L
10 W
–j 20 W
Fig. Q.8
r
100W
25 W
V
L
r
C
Q
rr
Fig. Q.9
8.9
VARIATION OF IMPEDANCE WITH FREQUENCY
The impedance of a parallel resonant circuit is maximum at the resonant
LO 5
frequency and decreases at lower
and higher frequencies as shown in
Fig. 8.15.
At very low frequencies, XL is
very small and XC is very large, so
the total impedance is essentially
inductive. As the frequency
increases, the impedance also
increases, and the inductive reactance dominates until the resonant
frequency is reached. At this point XL 5 XC, and the impedance is
at its maximum. As the frequency goes above resonance, capacitive
reactance dominates and the impedance decreases.
Fig. 8.15
302
8.10
Circuits and Networks
Q-FACTOR OF PARALLEL RESONANCE
LO 5
Consider the parallel RLC circuit shown in Fig. 8.16.
In the circuit shown, the condition for resonance occurs when the susceptance part is zero.
Admittance Y = G + jB
Fig. 8.16
=
1
1
+ j vC +
R
j vL
=
1
+
R

1 
j vC −


vL 
(8.22)
(8.23)
The frequency at which resonance occurs is
vr C −
1
=0
vr L
vr =
(8.24)
1
(8.25)
LC
The voltage and current variation with frequency is shown in Fig. 8.17. At resonant frequency, the current
is minimum.
The bandwidth,
BW 5 f2 – f1
For parallel circuit, to obtain the lower half-power frequency,
1
1
v1C −
=−
v1 L
R
From Eq. (8.26), we have
v
1
=0
v12 + 1 −
RC LC
(8.26)
(8.27)
Fig. 8.17
Resonance
303
If we simplify Eq. (8.27), we get
v1 =
−1
+
2 RC
 1 2
1

+
 2 RC 
LC
(8.28)
Similarly, to obtain the upper half-power frequency
v 2C −
1
1
=
v2 L R
(8.29)
From Eq. (8.29), we have
v2 =
Bandwidth
1
+
2 RC
 1 2
1

+
 2 RC 
LC
(8.30)
1
RC
BW = v2 − v1 =
The quality factor is defined as Qr =
Qr =
vr
v2 − v1
vr
= vr RC
1 / RC
In other words,
Q = 2p×
maximum energy stored
Energy dissipated /cycle
In the case of an inductor,
1
The maximum energy stored = LI 2
2
 I 2
Energy dissipated per cycle =   × R ×T
 2 
The quality factor Q = 2p×
∴ Q = 2p ×
1 / 2( LI 2 )
I2
1
R×
f
2
2
1  V 
L   R
2  vL 
2pf LR
V2 1
×
2
f
R
vL
v L
For a capacitor, maximum energy stored 5 1/2 (CV 2)
=
2 2
=
Energy dissipated per cycle = P ×T =
1
V2
×
2× R f
304
Circuits and Networks
The quality factor Q = 2p×
1 / 2(CV 2 )
V2 1
×
2R f
5 2pfCR 5 vCR
8.11
MAGNIFICATION
LO 5
Current magnification occurs in a parallel resonant circuit. The voltage applied to the parallel circuit, V 5 IR
Since
IL =
V
IR
=
= IQr
vr L vr L
For the capacitor, I C =
V
= IRvr C = IQr
1 / vr C
Therefore, the quality factor Qr 5 IL/I or IC/I
8.12
REACTANCE CURVES IN PARALLEL RESONANCE
LO 5
The effect of variation of frequency on the reactance of the parallel circuit is shown in Fig. 8.18.
Fig. 8.18
The effect of inductive susceptance,
−1
BL =
2pf L
Inductive susceptance is inversely proportional to the frequency or v. Hence, it is represented by a rectangular
hyperbola, MN. It is drawn in fourth quadrant, since BL is negative. Capacitive susceptance, BC 5 2pfC. It is directly
proportional to the frequency f or v. Hence it is represented by OP, passing through the origin. Net susceptance
B 5 BC – BL. It is represented by the curve JK, which is a hyperbola. At the point vr, the total susceptance is zero,
and resonance takes place. The variation of the admittance Y and the current I is represented by curve VW. The
current will be minimum at resonant frequency.
305
Resonance
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 5
The
impedance
Z1 5 (5 1 j3) V
and
Z2 5 (10 – j30) V are connected in parallel as
shown in Fig. Q.1. Find the value of X3 which
will produce resonance at the terminals a and b.
rrr 8-5.2 A current source is applied to the parallel
arrangement of R, L, and C where R 5 12 V,
L 5 2 H and C 5 3 mF. Compute the resonant
frequency in rad/s. Find the quality factor.
Fig. Q.1
Calculate the value of bandwidth. Compute the
lower and upper frequency of the bandwidth. Compute the voltage appearing across the parallel
elements when the input signal is i(t) 510 sin 1800 t.
rrr 8-5.3 For the circuit shown in Fig. Q.3, determine the value of RC for which the given circuit resonates.
rrr8-5.4 Using PSpice, obtain the frequency response of the circuit shown in Fig. Q.4. Use a linear
frequency sweep. Consider 1 , f , 1000 with 100 points.
rrr 8-5.1
Fig. Q.3
rrr 8-5.5
Fig. Q.4
Using PSpice, obtain bodeplots for V0 over a frequency from 1 kHz to 100 kHz for the circuit
shown in Fig. Q.5, using 20 points per decade.
Fig. Q.5
rrr 8-5.6
In a parallel resonance circuit shown in Fig. Q.6
find the (a) resonance frequency, (b) dynamic
resistance, and (c) bandwidth.
Fig. Q.6
Frequently Asked Questions linked to LO 5
rrr
Q
306
Circuits and Networks
rrr
RL
rrr
RL
–j20 W
20 W
–j10 W
Fig. Q.3
rrr 8-5.4
8.13
A parallel resonant circuit is a current amplifier. Justify.
[PU 2010]
LOCUS DIAGRAMS
A phasor diagram may be drawn and is expanded to develop a curve; known as
LO 6
a locus. Locus diagrams are useful in determining the behaviour or response
of an RLC circuit when one of its parameters is varied while the frequency and
voltage kept constant. The magnitude and phase of the current vector in the
circuit depends upon the values of R, L, and C and frequency at the fixed source
voltage. The path traced by the terminus of the current vector when the parameters R, L, or C are varied while
f and v are kept constant is called the current locus.
The term circle diagram identifies locus plots that are either circular or semicircular loci of the terminus
(the tip of the arrow) of a current phasor or voltage phasor. Circle diagrams are often employed as aids in
analysing the operating characteristics of circuits like equivalent circuit of transmission lines and some types
of ac machines.
Locus diagrams can be also drawn for reactance, impedance, susceptance, and admittance when frequency is
variable. Loci of these parameters furnish important information for use in circuit analysis. Such plots are particularly
useful in the design of electric wave filters.
8.13.1 Series Circuits
To discuss the basis of representing a series circuit by means of a circle diagram, consider the circuit shown in Fig.
8.19 (a). The analytical procedure is greatly simplified by assuming that inductance elements have no resistance and
that capacitors have no leakage current.
The circuit under consideration has constant reactance but variable resistance.
The applied voltage will be assumed with constant rms voltage V. The power
V
factor angle is designated by u. If R 5 0, IL is obviously equal to
and has
XL
maximum value. Also, I lags V by 90°. This is shown in Fig. 8.19 (b). If R is
V
increased from zero value, the magnitude of I becomes less than
and u
XL
becomes less than 90° and finally when the limit is reached, i.e. when R equals
Fig. 8.19 (a)
to infinity, I equals to zero and u equals to zero. It is observed that the tip of the
current vector represents a semicircle as indicated in Fig. 8.19 (b).
In general,
V
IL =
Z
307
Resonance
X
Z
X
or Z = L
sin u
V
sin u
I=
(8.31)
XL
For constant V and X, Eq. (8.31) is the polar equation of a
sin u =
circle with diameter V . Figure 8.19 (b) shows the plot of Eq.
XL
(8.31) with respect to V as reference.
The active component of the current IL in Fig. 8.19 (b) is OIL cos
u which is proportional to the power consumed in the RL circuit. In a
similar way we can draw the loci of current if the inductive reactance is
replaced by a capacitive reactance as shown in Fig. 8.19 (c). The current
semicircle for the RC circuit with variable R will be on the left-hand side
V
of the voltage vector OV with diameter
as shown in Fig. 8.19 (d). The
XL
current vector OIC leads V
Fig. 8.19 (b)
Fig. 8.19 (c)
by u°. The active component of the current IC X in Fig. 8.19 (d)
is OIC cos u which is proportional to the power consumed in the
RC circuit.
Circle Equations for an RL Circuit
Fixed reactance and variable resistance. The X-co-ordinate and
Y-co-ordinate of IL in Fig. 8.19 (b) respectively are IX 5 IL sin u;
Iy 5 IL cos u,
Where
∴
X
V
R
; sin u = L ; cos u = ; Z = R 2 + X L2
Z
Z
Z
XL
V XL
=V ⋅ 2
IX = ⋅
Z Z
Z
R
V R
IY = ⋅ = V ⋅ 2
Z Z
Z
IL =
Fig. 8.19 (d)
(8.32)
(8.33)
Squaring and adding Eqs (8.32) and (8.33), we obtain
I X2 + IY2 =
V2
(8.34)
2
R + X L2
From Eq. (8.32),
Z 2 = R 2 + X L2 = V ⋅
∴ Eq. (8.34) can be written as
or
I X2 + IY2 −
V
⋅IX = 0
XL
XL
IX
I X2 + IY2 =
V
⋅IX
XL
308
Circuits and Networks
 V
Adding 
 2 X
2
 to both sides the above equation can be written as

L
2
2


 I − V  + I 2 =  V 
Y
 2 X L 
 X 2 X L 
(8.35)
Equation (8.35) represents a circle whose radius is
V
V
and the coordinates of the centre are
, 0.
2XL
2XL
In a similar way, we can prove that for a series RC circuit as in Fig. 8.19 (c), with variable R, the locus
of the tip of the current vector is a semicircle and is given by
2

2
 I + V  + I 2 = V

X
Y

2 X C 
4 X C2
The centre has co-ordinates of −
(8.36)
V
V
, 0 and radius as
.
2XL
2XL
Fixed Resistance, Variable Reactance
Consider the series RL circuit with constant resistance R but
variable reactance XL as shown in Fig. 8.20 (a).
V
and u 5 0, the power factor of the circuit becomes unity;
R
as the value XL is increased from zero, IL is reduced and finally when XL is a, current becomes zero and u will
be lagging behind the voltage by 90° as shown in Fig. 8.20 (b). The current vector describes a semicircle with
diameter V and lies in the right-hand side of voltage vector OV. The active component of the current OIL cos u
R
is proportional to the power consumed in the RL circuit.
When XL 5 0; IL assumes maximum value of
(a)
Fig. 8.20
(b)
Equation of Circle
Consider Eq. (8.34)
I X2 + IY2 =
From Eq. (8.33)
V2
R 2 + X L2
Z 2 = R 2 + X L2 =
Substituting Eq. (8.37) in Eq. (8.34),
V
I X2 + IY2 = IY
R
VR
IY
(8.37)
(8.38)
309
Resonance
I X2 + IY2 −
V
IY = 0
R
 V 2
Adding   to both sides the above equation can be written as
 2 R 
2
 V 2

V 
I X2 +  IY −  =  
 2 R 

2R 
(8.39)
V
V
and the coordinates of the centre are 0;
.
2R
2R
Let the inductive reactance in Fig. 8.20 (a) be replaced by a capacitive reactance as shown in Fig. 8.21 (a).
The current semicircle of a RC circuit with variable XC will be on the left-hand side of the voltage vector
V
OV with diameter . The current vector OIC leads V by u°. As before, it may be proved that the equation of
R
the circle shown in Fig. 8.21 (b) is
Equation (8.39) represents a circle whose radius is
2
 V 2

V 
I X2 +  IY −  =  
 2 R 

2R 
(a)
Fig. 8.21
(b)
EXAMPLE 8.9
For the circuit shown in Fig. 8.22 (a), plot the locus of the current, mark the
range of I for maximum and minimum values of R, and the maximum power
consumed in the circuit. Assume XL 5 25 V and R 5 50 V. The voltage is
200 V; 50 Hz.
Solution Maximum value of current
200
= 8 A; u = 90°
25
Minimum value of current
200
I min =
= 3.777 A; u = 27.76°
2
2
(50) + (25)
I max =
Fig. 8.22 (a)
310
Circuits and Networks
The locus of the current is shown in Fig. 8.22 (b).
Power consumed in the circuit is proportional to I cos u for
constant V. The maximum ordinate possible in the semicircle (AB
in Fig. 8.22 (b)) represents the maximum power consumed in the
circuit. This is possible when u 5 45°, under the condition power
1
factor cos u 5 cos 45° 5
.
2
Hence, the maximum power consumed in the circuit
I
V × AB = V × max
L
V
I max =
= 84 A
XL
Fig. 8.22 (b)
2
Pmax
(200)
V2
=
=
= 800 W
2XL
2× 25
EXAMPLE 8.10
For the circuit shown in Fig. 8.22 (a), if the reactance is variable, plot the range of I for maximum and
minimum values of XL and maximum power consumed in the circuit.
Solution
Maximum value of current I max =
Minimum value of current I min =
200
= 4 A; u = 0
50
200
2
2
(50) + (25)
= 3.777 A; u = 27.76°
The locus of current is shown in Fig. 8.23.
Maximum power will be when I 5 4 A
Hence, Pmax 5 4 3 200 5 800 W
Fig. 8.23
EXAMPLE 8.11
For the circuit shown in Fig. 8.24 (a), draw the locus of the current. Mark the
range of I for maximum and minimum values. Assume XC 5 50 V; R 5 10 V;
V 5 400 V.
Solution
I max =
I min =
400
= 40 A; u = 0°
10
400
2
2
(50) + (10)
Fig. 8.24 (a)
= 7.716 A.u = tan−1 5 = 77.8°
The locus of the current is shown in Fig. 8.24 (b).
Fig. 8.24 (b)
Resonance
311
8.13.2 Parallel Circuits
Variable XL Locus plots are drawn for parallel branches containing RLC elements in a similar way
as for series circuits. Here we have more than one current locus. Consider the parallel circuit shown in Fig.
8.25 (a). The quantities that may be varied are XL, XC, RL and RC for a given voltage and frequency.
Let us consider the variation of XL from zero to `. Let OV shown in
Fig. 8.25 (b), be the voltage vector, taken as reference. A current, IC, will flow
in the condenser branch whose parameters are held constant and leads V by an
X 
angle uC = tan−1  C  , when XL 5 0, the current IL, through the inductive
 RC 
V
branch is maximum and is given by
and it is in phase with the applied
RL
voltage. When XL is increased from zero, the current through the inductive
−1 X L
branch IL decreases and lags V by uL = tan
as shown in Fig. 8.25 (b).
RL
Fig. 8.25 (a)
For any value of IL, the IL RL drop and IL XL drop must add at right angles to
give the applied voltage. These drops are shown as OA and AV respectively. The locus of IL is a semicircle, and
the locus of ILRL drop is also a semicircle. When XL 5 0, i.e. IL is maximum, IL coincides with the diameter of
its semicircle and ILRL drop also coincides with the diameter of its semi-circle as shown in the figure; both these
semicircles are shown with dotted circles as OILB and OAV respectively.
Since the total current is IC 1 IL. For example, for a particular value of IC and IL, the total current is represented by
OC on the total current semicircle. As XL is varied, the locus of the resultant current is therefore, the circle IC CB as
shown with thick line in the Fig. 8.25 (b).
Fig. 8.25 (b)
Variable XC A similar procedure can be adopted as outlined above to draw the locus plots of Il and I
when XC is varying while RL, RC, XL, V and f are held constant. The curves are shown in Fig. 8.25 (c).
OV presents the voltage vector, OB is the maximum current through RC branch when XL 5 0; OIL is the
C
current through the RL branch lagging OV by an angle uL = tan−1 L . As XC is increased from zero, the
RL
X
current through the capacitive branch IC decreases and leads V by uC = tan−1 C . For a particular IC, the
RC
312
Circuits and Networks
resultant current I 5 IL 1 IC and is given by OC. The dotted semicircle OICB is the locus of the IC, thick circle
ILCB is the locus of the resultant current.
Fig. 8.25 (c)
Variable RL The locus of current for the variation of RL in Fig. 8.26 (a) is shown in Fig. 8.26 (b). OV
represents the reference voltage, OILB represents the locus of IL and IC CB represents the resultant current
V
locus. Maximum I L =
is represented by OB.
XL
(a)
Fig. 8.26
(b)
Variable RC The locus of currents for the variation of RC in Fig. 8.27 (a) is plotted in Fig. 8.27 (b)
where OV is the source voltage and semicircle OAB represents the locus of IC. The resultant current locus is
given by BCIL.
313
Resonance
(a)
(b)
Fig. 8.27
EXAMPLE 8.12
For the parallel circuit shown in Fig. 8.28 (a), draw the locus
of I1 and I. Mark the range of values for R1 between 10 V and
100 V. Assume XL 5 25 V and R2 5 25 V. The supply voltage
is 200 V and frequency is 50 Hz, both held constant.
Fig. 8.28 (a)
Solution Let us take voltage as reference; on the positive X-axis. I2 is given by I 2 =
phase with V.
When
R1 = 10 V
I1 =
When
R2 = 100 V
I1 =
200
(100 + 625)
= 7.42 A; u1 = tan−1
200
(10000 + 625)
25
= 68.19°
10
= 1.94 A; u2 = tan−1
The variation of I1 and I are shown in Fig. 8.28 (b).
25
= 14.0°
100
200
= 8 A and is in
25
314
Circuits and Networks
Fig. 8.28 (b)
EXAMPLE 8.13
Draw the locus of I2 and I for the parallel circuit shown
in Fig. 8.29 (a).
Fig. 8.29 (a)
Solution I1
−1
leads the voltage by a fixed angle u1 given by tan
XC
.
R1
I2 varies according to the value of XC2.
Fig. 8.29 (b)
I2 is maximum when XC2 5 0 and is in phase with V.
I2 is zero when XC2 5 ` as shown in Fig. 8.29 (b).
EXAMPLE 8.14
For a parallel circuit shown in Fig. 8.30 (a), plot the locus of
currents.
Fig. 8.30 (a)
315
Resonance
Solution Current I1 leads the voltage by a fixed angle u1 given by
XC
, current I2 leads the voltage by 90°. I3 varies according to
R1
the value of XL, when XL 5 0, I3 is maximum and is given by V ; is
RL
in phase with V; when XL 5 `, I3 is zero. Both these extremities are
shown in Fig. 8.30 (b). For a particular value of I3 the total current I
is given by I1 1 I2 1 I3 5 OA 1 AB 1 BC.
tan−1
Fig. 8.30 (b)
Additional Solved Problems
PROBLEM 8.1
For the circuit shown in Fig. 8.31, determine the frequency at which
the circuit resonates. Also find the voltage across the inductor at
resonance and the Q-factor of the circuit.
Solution The frequency of resonance occurs when XL 5 XC
vL =
1
vC
∴ v=
=
1
LC
Fig. 8.31
rad/sec
1
= 447.2 rad/sec
0.1×50 ×10−6
1
fr =
(447.2) = 71.17 Hz
2p
The current passing through the circuit at resonance,
I=
V 100
=
= 10 A
R
10
The voltage drop across the inductor
VL 5 IXL 5 IvL
5 10 3 447.2 3 0.1 5 447.2 V
The quality factor
Q=
vL 447.2× 0.1
=
= 4.47
R
10
316
Circuits and Networks
PROBLEM 8.2
A series RLC circuit has a quality factor of 5 at 50 rad/s. The current flowing through the circuit at resonance
is 10 A and the supply voltage is 100 V. The total impedance of the circuit is 20 V. Find the circuit constants.
Solution The quality factor Q 5 5
At resonance, the impedance becomes resistance.
V
R
100
10 =
R
R = 10 V
vL
Q=
R
Q = 5, R = 10
vL = 50
50
L=
=1 H
v
1
Q=
vCR
1
C=
Q R
1
=
5×50 ×10
C = 400 mF
The current at resonance is I =
∴
Since
∴
∴
PROBLEM 8.3
A voltage v(t) 5 10 sin vt is applied to a series RLC circuit. At the resonant frequency of the circuit, the
maximum voltage across the capacitor is found to be 500 V. Moreover, the bandwidth is known to be 400
rad/s and the impedance at resonance is 100 V. Find the resonant frequency. Also find the values of L and C
of the circuit.
Solution The applied voltage to the circuit is
Vmax = 10 V
10
Vrms =
= 7.07 V
2
The voltage across capacitor VC 5 500 V
VC
500
=
= 70.7
V
7.07
The bandwidth BW 5 400 rad/s
The magnification factor Q =
v2 – v1 5 400 rad/s
317
Resonance
The impedance at resonance Z 5 R 5 100 V
vr
Since Q =
v2 − v1
vr 5 Q (v2 – v1) 5 28280 rad/s
28280
= 4499 Hz
2p
R
The bandwidth v2 − v1 =
L
fr =
∴
Since
L=
fr =
C=
100
R
=
= 0.25 H
v2 − v1 400
1
2p LC
1
2
(2pf r ) × L
=
1
2
2p×(4499) × 0.25
= 5 nF
PROBLEM 8.4
Find the value of L at which the circuit resonates at a frequency of 1000
rad/s in the circuit shown in Fig. 8.32.
Solution
Y=
Y=
1
1
+
10 − j12 5 + jX L
10 + j12
10 2 + 122
+
5 − jX L
Fig. 8.32
25 + X L2
 12
X L 
+ j  2
−
2
10 + 12
25 +
25 + X L2 
10 + 12
At resonance, the susceptance becomes zero.
XL
12
= 2
Then
2
25 + X L 10 + 122
=
10
2
2
+
5
X L2
12XL2 – 244 XL 1 300 5 0
From the above equation,
XL2 – 20.3 XL 1 25 5 0
XL =
+20.3 ±
2
(20.3) − 4 × 25
2
20.3 + 412 −100
or
2
5 18.98 V or 1.32 V
∴ XL 5 vL 5 18.98 or 1.32 V
=
20.3 − 412 −100
2
318
Circuits and Networks
L=
18.98
1.32
or
1000
1000
L 5 18.98 mH or 1.32 mH
PROBLEM 8.5
Two impedances Z1 5 20 1 j10 and Z2 5 10 – j30 are connected in parallel and this combination is connected
in series with Z3 5 30 1 jX. Find the value of X which will produce resonance.
Solution Total impedance is
Z = Z3 + ( Z1 || Z 2 )
(20 + j10)(10 − j 30)



= (30 + jX ) + 




20
10
10
30
j
j
+
+
−




200 − j 600 + j100 + 300
= (30 + jX ) +
30 − j 20
 500 − j 500 

= 30 + jX + 
 30 − j 20 
 500 1− j 30 + j 20 
(
)(
)

= 30 + jX + 

2
2


(30) + (20)


 500 (30 + j 20 − j 30 + 20) 

= (30 + jX ) + 

900 + 400


5
= 30 + jX + (50 − j10)
13




5
5
= 30 + ×50
0 + j  X − ×10





13
13
At resonance, the imaginary part is zero
∴
X−
50
=0
13
50
X=
= 3.85 V
13
PROBLEM 8.6
A 50 V resistor is connected in series with an inductor having
internal resistance, a capacitor and 100 V variable frequency
supply as shown in Fig. 8.33. At a frequency of 200 Hz, a maximum
current of 0.7 A flows through the circuit and voltage across the
capacitor is 200 V. Determine the circuit constants.
Solution At resonance, current in the circuit is maximum.
I 5 0.7 A
Fig. 8.33
319
Resonance
Voltage across capacitor is VC 5 IXC
Since VC 5 200, I 5 0.7
1
XC =
vC
0.7
vC =
200
0.7
∴ C=
= 2.785 mF
200 × 2p× 200
At resonance,
X L − XC = 0
∴
X L = XC
1
200
=
= 285.7 V
vC
0.7
X L = vL = 285.7 V
XC =
Since
285.7
= 0.23 H
2p× 200
At resonance, the total impedance
Z = R + 50
∴
∴
L=
R + 50 =
V 100
=
0.7
I
R + 50 = 142.86 V
∴
R = 92.86 V
PROBLEM 8.7
In the circuit shown in Fig. 8.34, a maximum current of 0.1 A flows
through the circuit when the capacitor is at 5 µF with a fixed
frequency and a voltage of 5 V. Determine the frequency at which
the circuit resonates, the bandwidth, the quality factor Q and the
value of resistance at resonant frequency.
Solution At resonance, the current is maximum in the circuits.
V
I=
R
V
5
∴ R= =
= 50 V
I
0.1
The resonant frequency is
1
vr =
LC
1
=
= 1414.2 rad/sec
0.1×5×10−6
fr =
1414.2
= 225 Hz
2p
Fig. 8.34
320
Circuits and Networks
The quality factor is Q =
vL 1414.2× 0.1
=
= 28
50
R
fr
=Q
BW
f
225
The bandwidth BW = r =
= 80.36 Hz
Q
2.8
Since
PROBLEM 8.8
In the circuit shown in Fig. 8.35, determine the circuit constants
when the circuit draws a maximum current at 10 µF with a 10 V, 100
Hz supply. When the capacitance is changed to 12 µF, the current
that flows through the circuit becomes 0.707 times its maximum
value. Determine Q of the coil at 900 rad/sec. Also find the maximum
current that flows through the circuit.
Solution At resonant frequency, the circuit draws maximum
Fig. 8.35
current. So, the resonant frequency fr 5 100 Hz
1
fr =
2p LC
1
L=
2
C ×(2pf r )
1
=
= 0.25 H
2
−6
10 ×10 (2p×100)
1
We have vL −
=R
vC
1
900 × 0.25 −
=R
900 ×12×10−6
∴ R = 132.4 V
vL 900 × 0.25
=
= 1.69
R
132.4
10
The maximum current in the circuit is I =
= 0.075 A
132.4
The quality factor Q =
PROBLEM 8.9
In the circuit shown in Fig. 8.36, the current is at its
maximum value with capacitor value C 5 20 µF and
0.707 times its maximum value with C 5 30 µF. Find the
value of Q at v 5 500 rad/s, and circuit constants.
Fig. 8.36
321
Resonance
Solution The voltage applied to the circuit is V 5 20 V. At resonance, the current in the circuit is maximum.
The resonant frequency vr 5 500 rad/s.
1
Since vr =
∴
XL =
Since we have vL −
LC
1
v2r C
=
1
2
(500) × 20 ×10−6
= 0.2 H
1
=R
vC
1
=R
500 ×30 ×10−6
∴ R = 100 − 66.6 = 33.4
vL 500 × 0.2
The quality factor is Q =
=
= 2.99
R
33.4
500 × 0.2 −
PROBLEM 8.10
In the circuit shown in Fig. 8.37, an inductance of 0.1 H having a Q of 5 is
in parallel with a capacitor. Determine the value of capacitance and coil
resistance at resonant frequency of 500 rad/s.
v L
Solution The quality factor Q = r
R
Since L = 0.1 H, Q = 5 and vr = 500 rad /s
Fig. 8.37
500 × 0.1
R
500 × 0.1
∴ R=
= 10 V
5
1
Since v2r =
LC
1
2
(500) =
0.1×C
Q=
∴ The capacitance value C =
1
2
0.1×(500)
= 40 mF
PROBLEM 8.11
A series RLC circuit consists of a 50 V resistance, 0.2 H inductance, and 10 µF capacitor with an applied
voltage of 20 V. Determine the resonant frequency. Find the Q-factor of the circuit. Compute the lower and
upper frequency limits and also find the bandwidth of the circuit.
Solution Resonant frequency
fr =
Quality factor Q =
1
2p LC
=
1
2p 0.2×10 ×10−6
vL 2p×112.5× 0.2
=
= 2.83
R
50
= 112.5 Hz
322
Circuits and Networks
Lower frequency limit
R
50
f1 = f r −
= 92.6 Hz
= 112.5 −
4 pL
4 × p× 0.2
Upper frequency limit
R
50
f2 = fr +
= 112.5 + 19.89 = 132.39 Hz
= 112.5 +
4 pL
4p× 0.2
Bandwidth of the circuit
BW 5 f2 – f1 5 132.39 – 92.6 5 39.79 Hz
PROBLEM 8.12
A tumed circuit consists of a coil having an inductance of 200 µH and a resistance of 15 V in parallel with a
series combination of a variable capacitance and resistor of 80 V. It is supplied by a 60 V source. If the supply
frequency is 1 MHz. what is the value of C to give resonance?
Fig. 8.38
Solution Total admittance of the circuit y =
Rationalising the above equation,
1
+
RL + jvL
1
Rc −
j
vc
Rc + j / vc
1
RL + v L
Rc2 + 2 2
v c
Separating the real and imaginary parts,
y=
=
RL − jvL
2
2 2
RL
RL2
2 2
+v L
+
+
 1 / vc
vL 

+ j
− 2
2 
 2
1
1
2
v
R
L
+
L
Rc + 2 2
 Rc + 2 2



v c
v c
Rc
At resonance, susceptance part 5 0
v = vr
1
vr L
vr C
=
Hence
1
2
RL2 + v2r L2
RC + 2 2
vr C
323
Resonance

1 
1  2
2 2
vr L  RC2 + 2 2  =
 RL + vr L 


v
c
vr c 
r


1  2
1 
2 2
vr2  RC2 + 2 2  =
 RL + vr L 



vr c  LC

RL2 − L / C
1
vr =
LC RC2 − L / C
1
LC
=
1
RL2 − L / C
LC
RC2
−L /C

∵ vr =

1 

LC 
L
L
= RL2 −
C
C
R C = RL
RC2 −


1
(15 + j1256) ⋅ 80 − j

6

2pC ×10 
Imaginary part 5 0 at resonance
15
1256 ×80 =
2pC ×106
C = 23.76 PF
PROBLEM 8.13
For the parallel circuit shown in Fig. 8.39; V 5 200 V; R2 5 50 V; X1 5 25 V . R1 is varied from 10 V to
50 V, draw the locus diagram. Find maximum and minimum values of source current.
Fig. 8.39
Solution
I = I1 + I 2 ; I 2 =
I2 =
When
V
; I1 =
R2
V
X 
R12 + X12 tan−1  1 
 R1 
200 0°
=4A
50
R1 = 10 V ; I1 =
200 0°
100 + 625 tan

−1 
 25 
 
10
= 7.42 −68.2°
324
Circuits and Networks
Wheen
R1 = 50 V ; I1 =
200 0°
2500 + 625 tan
Maximum value of

−1 
 25 
= 3.57 −26.6°
 
50
I = 4 + 7.42 −68.2°
= 4 + 2.76 − j 6.9
= 6.76 − j 6.9 = 9.66 −45.58°
Minimum value of
I = 4 + 3.57 −26.6°
= 4 + 3.192 − j1.6
= 7.192 − j1.6 = 7.36 −12.54°
The locus diagram is shown in Fig. 8.40.
Fig. 8.40
PROBLEM 8.14
A series circuit has a resonance frequency of 120 kHz, a bandwidth of 50 kHz, and Q52. Determine the cutoff frequencies.
Solution Since Q < 10, we cannot use the formulas f 2 = f r +
f1 = f r −
B ⋅W
2
B ⋅W = f 2 − f1
50, 000 = f 2 − f1
f 2 = 50, 000 + f1
fr =
also
,
f1 f 2
B ⋅W
2
325
Resonance
1, 20, 000 =
f1 f 2
14.4 ×109 = f1 f 2
14.4 ×109 = f1 (50, 000 + f1 )
f12 + 50, 000 f1 −14.4 ×109 = 0
−50, 000 ± (50, 000) + 4 (14.4 ×109 )
2
f1 =
2
−50, 000 ± 24.5×10 4
2
f1 = −147.5 kHz ; 195 kHz
=
Since 2ve frequency does not exist,
f1 = 195 kHz
50 = f 2 −195 ⇒ f 2 = 245 kHz
PROBLEM 8.15
A series RLC circuit is supplied at 220 V :
50 Hz. At resonance, the voltage across the
capacitor 5 550 V, I 5 1 A. Determine R, L, and C.
VC
Solution At resonance, XL 5 Xc
Current at resonance,
I=
V
V
220
= =
R + j ( X L − XC ) R
R
220
∴ I=
⇒ R = 220 Ω
R
VC = I o X C
1
550 = 1⋅
voC
C=
fo =
1
⋅ 5.78 mF
550 × 2p×50
1
2p LC
2
 1 2
1  1 
LC = 

 ⇒ L = 
 2pf o 
C  2pf o 
L=
 1 2

 = 1.75 H
5.78×10−6 100p 
1
Fig. 8.41
326
Circuits and Networks
Hence, the elements are R 5 220 V
L 5 1.75 H
C 5 5.78 mF
PSpice Problems
PROBLEM 8.1
Using PSpice, for the circuit shown in Fig 8.42, determine the
frequency at which the circuit resonates. Also find the voltage
across the inductor at resonance and Q-factor of the circuit.
*
TO OBTAIN RESONANT FREQUENCY
VS 1
0
AC
100
0
R
1
2
10
L
2
3
0.1
Fig. 8.42
C
3
0
50 U
.AC LIN 1000 1 100
.PROBE I(L)
.PLOT AC VM(L) VP(L)
.END
*** AC ANALYSIS
TEMPERATURE 5 27.000 DEG C
***************************************************************
LEGEND:
* : VM(L)
1: VP(L)
FREQ
VM(L)
(*)------ 1.0000E–02 1.0000E100 1.0000E102 1.0000E104 1.0000E106
(1)------ 0.0000E100 5.0000E101 1.0000E102 1.5000E102 2.0000E102
___________________________
1.000E 1 00 1.974E – 02 .
*
.
.
.
1
.
1.099E 1 00 2.385E – 02 .
*
.
.
.
1
.
1.198E 1 00 2.835E – 02 .
*
.
.
.
1
.
1.297E 1 00 3.323E – 02 .
*
.
.
.
1
.
1.396E 1 00 3.850E – 02 .
*
.
.
.
1
.
1.496E 1 00 4.417E – 02 .
*
.
.
.
1
.
1.595E 1 00 5.022E – 02 .
*
.
.
.
1
.
1.694E 1 00 5.666E – 02 .
*
.
.
.
1
.
1.793E 1 00 6.348E – 02 .
*
.
.
.
1
.
9.960E 1 01 1.943E 1 02 .
1
.
.*
.
.
9.970E 1 01 1.939E 1 02 .
1
.
.*
.
.
327
Fig. 8.43
Resonance
9.980E 1 01 1.936E 1 02 .
1
.
.*
9.990E 1 01 1.932E 1 02 .
1
.
.*
1.000E 1 02 1.929E 1 02 .
1
.
.*
1.001E 1 02 1.926E 1 02 .
1
.
.*
--------------------------JOB CONCLUDED
.
.
.
.
Result
Q- Factor 5 vr /BW 5 71.618/15.878 5 4.51, Refer Fig. 8.43.
.
.
.
.
328
Circuits and Networks
PROBLEM 8.2
For circuit shown in Fig. 8.44, the applied voltage is
v(t) 5 15 sin1800 t. Determine the resonant frequency.
Calculate the quality factor and bandwidth. Compute the
lower and upper limits of the bandwidth.
1800
v
=
2p
2p
15
=
= 10.607 V
2
f =
Vrms
Fig. 8.44
*
VS 1
0
AC 10.607 0
R
1
2
15
L
2
3
5
C
3
0
2U
.AC LIN 100 40 60
.PROBE I(VS)
.PLOT AC IM(VS) IP(VS)
.END
**** AC ANALYSIS TEMPERATURE 5 27.000 DEG C
***************************************************************
LEGEND:
*: IM(VS)
1: IP(VS)
FREQ IM(VS)
(*)----- 1.0000E–02 1.0000E–01 1.0000E100 1.0000E101 1.0000E102
(1)----- –2.0000E102 –1.0000E102 0.0000E100 1.0000E102 2.0000E102
___________________________
4.000E 1 01 1.447E – 02 . * . 1
.
.
.
4.020E 1 01 1.480E – 02 . * . 1
.
.
.
4.040E 1 01 1.514E – 02 . * . 1
.
.
.
4.061E 1 01 1.550E – 02 . * . 1
.
.
.
4.869E 1 01 1.001E – 01 .
1*
.
.
.
4.889E 1 01 1.140E – 01 .
1*
.
.
.
4.909E 1 01 1.323E – 01 .
1* .
.
.
4.929E 1 01 1.572E – 01 .
1* .
.
.
4.950E 1 01 1.930E – 01 .
1. * .
.
.
4.970E 1 01 2.484E – 01 .
1. * .
.
.
4.990E 1 01 3.420E – 01 .
1. * .
.
.
5.010E 1 01 5.106E – 01 . 1 . * .
.
.
5.030E 1 01 7.029E – 01 . 1
. *.
.
.
5.051E 1 01 5.697E – 01 .
. *.
. 1 .
5.071E 1 01 3.787E – 01 .
. * .
. 1
.
329
Fig. 8.45
Resonance
5.091E 1 01
5.111E 1 01
5.131E 1 01
5.152E 1 01
5.172E 1 01
5.192E 1 01
5.212E 1 01
2.705E – 01 .
2.079E – 01 .
1.683E – 01 .
1.411E – 01 .
1.215E – 01 .
1.066E – 01 .
9.498E – 02 .
. *
. *
.*
.*
.*
.*
*
Result
Q-FACTOR 5 105.35, Refer Fig. 8.45.
.
.
.
.
.
.
.
.1
.1
1
1
1
1
1
.
.
.
.
.
.
.
330
Circuits and Networks
Answers to Practice Problems
8-3.1
8-3.2
8-3.6
8-3.7
875.35 Hz; 914.42 Hz; 836.28 Hz; 0.2 H;
0.165 mF
50.3 Hz; 63.2 V; 3 (approx.)
Q 5 1; R 5 60 V; C 5 50 mF
100 V
8-3.8
f  L
1
f c = r 
− 
2
Q  CR
2
where Q =
8-3.10
1 L
; and
R C
fr =
1
2p LC
(i) 100 V (ii) 1002.5 Hz
8-5.1
2.07 V
8-5.3
1.77
8-5.6
(i) fr 5 1559.4 Hz (ii) 50 V (iii) 318.31 Hz
Objective-Type Questions
rrr 8.1
rrr 8.2
rrr 8.3
rrr 8.4
rrr 8.5
rrr 8.6
rrr 8.7
rrr 8.8
rrr 8.9
rrr 8.10
rrr 8.11
rrr 8.12
What is the total reactance of a series RLC circuit at resonance?
(a) equal to XL
(b) equal to XC
(c) equal to R
(d) zero
What is the phase angle of a series RLC circuit at resonance?
(a) zero
(b) 90°
(c) 45°
(d) 30°
In a series circuit of L 5 15 mH, C 5 0.015 mF, and R 5 80 V, what is the impedance at the resonant
frequency?
(a) (15 mH) v
(c) 80 V
(b) (0.015 F) v
(d) 1/(v 3 (0.015))
In a series RLC circuit operating below the resonant frequency, the current
(a) I leads VS
(b) I lags behind VS
(c) I is in phase with VS
In a series RLC circuit, if C is increased, what happens to the resonant frequency?
(a) It increases
(c) It remains the same (b) It decreases
(d) It is zero
In a certain series resonant circuit, VC 5 150 V, VL 5 150 V, and VR 5 50 V. What is the value of the source
voltage?
(a) Zero
(b) 50 V
(c) 150 V
(d) 200 V
A certain series resonant circuit has a bandwidth of 1000 Hz. If the existing coil is replaced by a coil with a
lower Q, what happens to the bandwidth?
(a) It increases
(c) It is zero
(b) It decreases
(d) It remains the same
In a parallel resonance circuit, why does the current lag behind the source voltage at frequencies below
resonance?
(a) Because the circuit is predominantly resistive
(b) Because the circuit is predominantly inductive
(c) Because the circuit is predominantly capacitive (d) None of the above
In order to tune a parallel resonant circuit to a lower frequency, the capacitance must
(a) be increased
(c) be zero
(b) be decreased
(d) remain the same
What is the impedance of an ideal parallel resonant circuit without resistance in either branch?
(a) Zero
(b) Inductive
(c) Capacitive
(d) I0nfinite
If the lower cut-off frequency is 2400 Hz and the upper cut-off frequency is 2800 Hz, what is the bandwidth?
(a) 400 Hz
(b) 2800 Hz
(c) 2400 Hz
(d) 5200 Hz
What values of L and C should be used in a tank circuit to obtain a resonant frequency of 8 kHz? The
bandwidth must be 800 Hz. The winding resistance of the coil is 10 V.
(a) 2 mH, 1 mF
(c) 1.99 mH, 0.2 mF
(b) 10 H, 0.2 mF
(d) 1.99 mH, 10 mF
For interactive quiz with answers,
scan the QR code given here
OR
visit
http://qrcode.flipick.com/index.php/266
9
LEARNING OBJECTIVES
9.1
POLYPHASE SYSTEM
In an ac system, it is possible to connect two or more individual circuits to a common polyphase source.
Though it is possible to have any number of sources in a polyphase system, the increase in the available power
is not significant beyond the three-phase system. The power generated by the same machine increases 41.4
per cent from single phase to two-phase, and the increase in the power is 50 per cent from single phase to
three-phase. Beyond three-phase, the maximum possible increase is only seven per cent, but the complications
are many. So, an increase beyond three-phase does not justify the extra complications. In view of this, it is
only in exceptional cases where more than three phases are used. Circuits supplied by six, twelve, and more
phases are used in high power radio transmitter stations. Two-phase systems are used to supply two-phase
servo motors in feedback control systems.
In general, a three-phase system of voltages (currents) is merely a combination of three single-phase
systems of voltages (currents) of which the three voltages (currents) differ in phase by 120 electrical degrees
from each other in a particular sequence. One such three-phase system of sinusoidal voltages is shown in
Fig. 9.1.
332
Circuits and Networks
Fig. 9.1
9.2
ADVANTAGES OF A THREE-PHASE SYSTEM
It is observed that the polyphase, especially three-phase, system has many advantages
LO 1
over the single-phase system, both from the utility point of view as well as from the
consumer point of view. Some of the advantages are as under.
1. The power in a single-phase circuit is pulsating. When the power factor of
the circuit is unity, the power becomes zero 100 times in a second in a 50 Hz
supply. Therefore, single-phase motors have a pulsating torque. Although the power supplied by each
phase is pulsating, the total three-phase power supplied to a balanced three-phase circuit is constant
at every instant of time. Because of this, three-phase motors have an absolutely uniform torque.
2. To transmit a given amount of power over a given length, a three-phase transmission circuit requires less
conductor material than a single-phase circuit.
3. In a given frame size, a three-phase motor or a three-phase generator produces more output than its
single-phase counterpart.
4. Three-phase motors are more easily started than single-phase motors. Single phase motors are not
self-starting, whereas three-phase motors are.
In general, we can conclude that the operating characteristics of a three-phase apparatus are superior
than those of a similar single-phase apparatus. All three-phase machines are superior in performance. Their
control equipment are smaller, cheaper, lighter in weight and more efficient. Therefore, the study of threephase circuits is of great importance.
Frequently Asked Questions linked to LO1*
rrr 9-1.1
9.3
What are the advantages of a three-phase system?
[AU May/June 2013]
GENERATION OF THREE-PHASE VOLTAGES
Three-phase voltages can be generated in a stationary armature with a rotating field
structure, or in a rotating armature with a stationary field as shown in Fig. 9.2 (a) and (b).
LO 2
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
Polyphase Circuits
333
Fig. 9.2
Single-phase voltages and currents are generated by single-phase generators as shown in Fig. 9.3 (a).
The armature (here a stationary armature) of such a generator has only one winding, or one set of coils. In
a two-phase generator, the armature has two distinct windings, or two sets of coils that are displaced 90°
(electrical degrees) apart, so that the generated voltages in the two phases have 90° phase displacement as
shown in Fig. 9.3 (b). Similarly, three-phase voltages are generated in three separate but identical sets of
windings or coils that are displaced by 120 electrical degrees in the armature, so that the voltages generated
in them are 120° apart in time phase. This arrangement is shown in Fig. 9.3 (c). Here, RR9 constitutes one
coil (R-phase); YY9 another coil (Y-phase), and BB9 constitutes the third phase (B-phase). The field magnets
are assumed in clockwise rotation.
The voltages generated by a three-phase alternator is shown in Fig. 9.3 (d). The three voltages are of
the same magnitude and frequency, but are displaced from one another by 120°. Assuming the voltages to
be sinusoidal, we can write the equations for the instantaneous values of the voltages of the three phases.
Counting the time from the instant when the voltage in phase R is zero.
The equations are
vRR9 5 Vm sin vt
vYY 9 5 Vm sin (vt – 120°)
vBB 9 5 Vm sin (vt – 240°)
At any given instant, the algebraic sum of the three voltages must be zero.
334
Circuits and Networks
Fig. 9.3
9.4
PHASE SEQUENCE
The sequence of voltages in the three phases are in the order vRR9 – vYY9 – vBB9, and
LO 3
they undergo changes one after the other in the above mentioned order. This is called
the phase sequence. It can be observed that this sequence depends on the rotation of
the field. If the field system is rotated in the anticlockwise direction, then the sequence of the voltages in the
three-phases are in the order vRR9 – vBB9 – vYY9; briefly we say that the sequence is RBY. Now the equations
can be written as
vRR9 5 Vm sin vt
vBB9 5 Vm sin (vt – 120°)
vYY9 5 Vm sin (vt – 240°)
EXAMPLE 9.1
What is the phase sequence of the voltages induced in the three
coils of an alternator shown in Fig. 9.4? Write the equations for
the three voltages.
Fig. 9.4
335
Polyphase Circuits
Solution Here, the field system is stationary and
the three coils, RR9, YY9 and BB9, are rotating in the
anticlockwise direction, so the sequence of voltages
is RBY, and the induced voltages are as shown in Fig.
9.4.
vRR9 5 Vm sin vt
vBB9 5 Vm sin (vt – 120°)
vYY9 5 Vm sin (vt – 240°) or Vm s in (vt 1 120°)
Fig. 9.5
EXAMPLE 9.2
What is the possible number of phase sequences in Fig. 9.4? What are they?
Solution There are only two possible phase sequences; they are RBY and RYB.
Frequently Asked Questions linked to LO3
rrr 9-3.1
9.5
What is phase sequence of a 3-phase system?
[AU May/June 2013]
INTERCONNECTION OF THREE-PHASE SOURCES AND LOADS
9.5.1 Interconnection of Three-phase Sources
In a three-phase alternator, there are three independent phase windings or coils.
LO 4
Each phase or coil has two terminals, viz. start and finish. The end connections
of the three sets of the coils may be brought out of the machine, to form three
separate single phase sources to feed three individual circuits as shown in
Fig. 9.6 (a) and (b).
The coils are interconnected to form a wye (Y ) or delta (D) connected three-phase system to achieve
economy and to reduce the number of conductors, and thereby, the complexity in the circuit. The threephase sources so obtained serve all the functions of the three separate single phase sources.
Fig. 9.6
9.5.2 Wye or Star-Connection
In this connection, similar ends (start or finish) of the three phases are joined together within the alternator as
shown in Fig. 9.7. The common terminal so formed is referred to as the neutral point (N), or neutral terminal.
Three lines are run from the other free ends (R, Y, B) to feed power to the three-phase load.
336
Circuits and Networks
Figure 9.7 represents a three-phase, four-wire, star-connected
system. The terminals R, Y, and B are called the line terminals of the
source. The voltage between any line and the neutral point is called
the phase voltage (VRN, VYN, and VBN), while the voltage between
any two lines is called the line voltage (VRY, VYB, and VBR). The
currents flowing through the phases are called the phase currents,
while those flowing in the lines are called the line currents. If the
neutral wire is not available for external connection, the system
is called a three-phase, three-wire, star-connected system. The
system so formed will supply equal line voltages displaced 120°
from one another and acting simultaneously in the circuit like
three independent single phase sources in the same frame of a
three-phase alternator.
Fig. 9.7
EXAMPLE 9.3
Figure 9.8 represents three phases of an alternator. Arrange the possible number
of three-phase star connections and indicate phase voltages and line voltages in
each case. (VRR9 5 VYY 9 5 VBB9)
Solution There are two possible star-connections and they can be arranged as
shown in Fig. 9.9 (a).
The phase voltages are
Fig. 9.8
VRN, VYN, VBN and VR9N, VY9N, VB9N
The line voltages are
VRY, VYB, VBR and VR9Y9, VY9B9, VB9R9
Note The phases can also be arranged as shown in Fig. 9.9 (b), in which case they do not look like a
star; so the name star or wye-connection is only a convention.
(a)
337
Polyphase Circuits
(b)
Fig. 9.9
9.5.3 Delta or Mesh-Connection
In this method of connection, the dissimilar ends of the windings are joined together, i.e. R9 is connected to
Y, Y 9 to B and B9 to R as shown in Fig. 9.10.
The three line conductors are taken from the three
junctions of the mesh or delta connection to feed the
three-phase load. This constitutes a three-phase, threewire, delta-connected system. Here there is no common
terminal; only three line voltages VRY, VYB and VBR are
available.
These line voltages are also referred to as phase voltages in
the delta-connected system. When the sources are connected
Fig. 9.10
in delta, loads can be connected only across the three line
terminals, R, Y, and B. In general, a three-phase source, star or delta, can be either balanced or unbalanced. A balanced
three-phase source is one in which the three individual sources have equal magnitude, with 120° phase difference as
shown in Fig. 9.3 (d).
EXAMPLE 9.4
Figure 9.11 represents three phases of an alternator. Arrange the possible number
of three-phase, delta connections and indicate phase voltages and line voltages in
each case (Note VRR9 5 VYY 9 5 VBB9).
Solution There are two possible delta connections which are shown as follows.
Vphase 5 Vline
The line voltages are
VRY, VYB, and VBR from Fig. 9.12 (a) and VRB, VBY, and VYR from Fig. 9.12 (b).
Fig. 9.11
338
Circuits and Networks
Fig. 9.12
9.5.4 Interconnection of Loads
The primary question in a star or delta-connected three-phase supply is how to apply the load to the threephase supply. An impedance, or load, connected across any two terminals of an active network (source) will
draw power from the source, and is called a single-phase load. Like alternator phase windings, a load can also
be connected across any two terminals, or between line and neutral terminal (if the source is star-connected).
Usually, the three-phase load impedances are connected in star or delta formation, and then connected to the
three-phase source as shown in Fig. 9.13.
Figure 9.13 (a) represents the typical interconnections of loads and sources in a three-phase star system,
and is of a three-phase four-wire system. A three-phase star- connected load is connected to a three-phase starconnected source, terminal to terminal, and both the neutrals are joined with a fourth wire. Figure 9.13 (b) is a
three-phase, three-wire system. A three-phase, delta-connected load is connected to a three-phase star-connected
source, terminal to terminal, as shown in Fig. 9.13 (b). When either source or load, or both are connected in delta,
only three wires will suffice to connect the load to source.
Fig. 9.13
Polyphase Circuits
339
Just as in the case of a three-phase source, a three-phase load can be either balanced or unbalanced. A
balanced three-phase load is one in which all the branches have identical impedances, i.e. each impedance
has the same magnitude and phase angle. The resistive and reactive components of each phase are equal.
Any load which does not satisfy the above requirements is said to be an unbalanced load.
EXAMPLE 9.5
Draw the interconnection between a three-phase, delta-connected source and a star-connected load.
Solution When either source or load, or both are connected in delta, only three wires are required to connect,
the load to source, and the system is said to be a three-phase, three-wire system. The connection diagram is
shown in Fig. 9.14.
Fig. 9.14
The three line voltages are VRY, VYB, and VBR.
EXAMPLE 9.6
Draw the interconnection between a three-phase, delta-connected source and delta-connected load.
Solution Since the source and load are connected in delta, it is a three-wire system. The connection diagram
is shown in Fig. 9.15.
Fig. 9.15
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 4
A three-phase, four-wire symmetrical 440 V; RYB system supplies a star-connected load in which
ZR 5 10 ∠0° V, ZY 5 10 ∠26.8° V and ZB 5 10 ∠–26.8° V. Find the line currents, the neutral
current and the load power.
rrr 9-4.2 Three impedances of (7 1 j4) V; (3 1 j2) V and (9 1 j2) V are connected between neutral and
the red, yellow and blue phases, respectively of a three-phase, four-wire system; the line voltage
is 440 V. Calculate (a) the current in each line, and (b) the current in the neutral wire.
rrr 9-4.1
340
Circuits and Networks
A symmetrical 3-phase, 3-wire, 440 V is connected to a star-connected load. The impedances in
each branch are Z1 5 (2 1 j3) V, Z2 5 (1– j2) V, Z3 5 (3– j4) V. Find its equivalent delta
connected load. Hence find the phase and line currents and the total power consumed in the
circuit.
rrr 9-4.4 Three impedances of (7 1 j4) V; (3 1 j2) V, and (9 1 j2) V are connected between neutral and
R,Y, and B phases. The line voltage of a 3-phase, four-wire system is 440 V; calculate the active
power in each phase and the total power drawn by the circuit.
rrr 9-4.3
Frequently Asked Questions linked to LO 4
rrr 9-4.1
9.6
Three inductive coils each having a resistance of 16 and a reactance of j 12 are connected in
star across a 400 V, 3 f, 50 Hz supply. Calculate phase voltage.
[AU Nov./Dec. 2012]
STAR-TO-DELTA AND DELTA-TO-STAR TRANSFORMATION
While dealing with currents and voltages in loads, it is often necessary to convert
LO 5
a star load to delta load, and vice versa. It has already been shown in Chapter 3
that delta (D) connection of resistances can be replaced by an equivalent star (Y )
connection and vice versa. Similar methods can be applied in the case of networks
containing general impedances in complex form. So also with ac, where the same
formulae hold good, except that resistances are replaced by the impedances. These formulae can be applied
even if the loads are unbalanced. Thus, considering Fig. 9.16 (a), star load can be replaced by an equivalent
delta-load with branch impedances as shown.
Delta impedances, in terms of star impedances, are
and
Z RY =
Z R ZY + ZY Z B + Z B Z R
ZB
ZYB =
Z R ZY + ZY Z B + Z B Z R
ZR
Z BR =
Z R ZY + ZY Z B + Z B Z R
ZY
Fig. 9.16
The converted network is shown in Fig. 9.16 (b). Similarly, we can replace the delta load of Fig. 9.16 (b)
by an equivalent star load with branch impedances as
ZR =
ZY =
and
ZB =
Z RY
Z RY Z BR
+ ZYB + Z BR
Z RY
Z RY ZYB
+ ZYB + Z BR
Z BR ZYB
Z RY + ZYB + Z BR
It should be noted that all impedances are to be expressed in their complex form.
341
Polyphase Circuits
EXAMPLE 9.7
A symmetrical three-phase, three-wire 440 V supply is connected
to a star-connected load as shown in Fig. 9.17 (a). The impedances
in each branch are ZR 5 (2 1 j3) V, ZY 5 (1 – j2) V and
ZB 5 (3 1 j4) V. Find its equivalent delta-connected load. The
phase sequence is RYB.
Solution The equivalent delta network is shown in Fig. 9.17 (b).
From Section 9.6, we can write the equations to find ZRY, ZYB,
and ZBR. We have
Z RY =
Z R ZY + ZY Z B + Z B Z R
ZB
Fig. 9.17 (a)
ZR 5 2 1 j3 5 3.61 ∠56.3°
ZY 5 1 – j2 5 2.23 ∠–63.4°
ZB 5 3 1 j4 5 5 ∠53.13°
ZRZY 1 ZYZB 1 ZBZR 5 (3.61 ∠56.3°)
(2.23 ∠–63.4°) 1 (2.23∠–63.4°)
(5∠53.13°) 1 (5∠53.13°) (3.61∠56.3°)
5 8.05 ∠–7.1° 1 11.15 ∠–10.27°
1 18.05 ∠109.43°
Fig. 9.17 (b)
Z RY =
5 12.95 1 j14.04 5 19.10 ∠47.3°
19.10 ∠47.3°
= 3.82 ∠− 5.83° = 3.8 − j 0.38
5 ∠53.13°
Z R ZY + ZY Z B + Z B Z R
ZR
19.10 ∠47.3°
=
= 5.29 ∠− 9° = 5.22 − j 0.82
3.61∠56.3°
ZYB =
Z BR =
=
Z R ZY + ZY Z B + Z B Z R
ZY
19.10 ∠47.3°
= 8.56 ∠110.7° = −3.02 + j8
2.23 ∠− 63.4°
The equivalent delta impedances are
ZRY 5 (3.8 – j0.38) V
ZYB 5 (5.22 – j0.82) V
ZBR 5 (– 3.02 1 j8) V
342
Circuits and Networks
EXAMPLE 9.8
A symmetrical three-phase, three-wire 400 V, supply is connected
to a delta-connected load as shown in Fig. 9.18 (a). Impedances
in each branch are ZRY 510 ∠30° V; ZYB 5 10 ∠–45° V and
ZBR 5 2.5 ∠60° V. Find its equivalent star-connected load; the
phase sequence is RYB.
Solution The equivalent star network is shown in Fig. 9.18 (b).
From Section 9.6, we can write the equations to find ZR, ZY,
and ZB as
ZR =
Z RY
Fig. 9.18 (a)
Z RY Z BR
+ ZYB + Z BR
ZRY 1 ZYB 1 ZBR 5 10 ∠30 1 10 ∠– 45 1 2.5 ∠60
5 (8.66 1 j5) 1 (7.07 – j7.07) 1 (1.25 1 j2.17)
5 16.98 1 j0.1 5 16.98 ∠0.33 V
ZR =
Fig. 9.18 (b)
(10 − 30)(2.5 − 60)
16.98 − 0.33
= 1.47 ∠89.67°
5 (0.008 1 j1.47) V
ZY =
=
ZB =
=
Z RY ZYB
Z RY + ZYB + Z BR
(10 − 30)(10 − 45)
= 5.89∠−15.33° V
16.98 − 0.33
Z BR ZYB
Z RY + ZYB + Z BRR
( 2.5 − 60)(10 − 45)
= 1.47∠14.67° V
16.98 − 0.33
The equivalent star impedances are
ZR 5 1.47 ∠89.67 V, ZY 5 5.89 ∠–15.33° V and ZB 5 1.47 ∠14.67 V
Balanced Star-Delta and Delta-Star Conversion
If the three-phase load is balanced, then the conversion formulae in Section 9.6 get simplified. Consider a
343
Polyphase Circuits
balanced star-connected load having an impedance Z1 in each phase as shown in Fig. 9.19 (a).
Fig. 9.19
Let the equivalent delta-connected load have an impedance of Z2 in each phase as shown in Fig. 9.19 (b).
Applying the conversion formulae from Section 9.6 for delta impedances in terms of star impedances, we
have
Z2 5 3Z1
Similarly, we can express star impedances in terms of delta as Z1 5 Z2/3.
EXAMPLE 9.9
Three identical impedances are connected in delta as shown in Fig.
9.20 (a). Find an equivalent star network such that the line current
is the same when connected to the same supply.
Fig. 9.20 (a)
Solution The equivalent star network is shown in Fig. 9.20 (b).
From Section 9.6.1, we can write the equation to find Z1 5 Z2/3
Z2 5 3 1 j4
5 5 ∠53.13 V
5
\ Z1 5
∠53.13
3
5 1.66 ∠53.15
5 (1.0 1 j1.33) V
Fig. 9.20 (b)
344
Circuits and Networks
Frequently Asked Questions linked to LO 5
rrr 9-5.1
rrr 9-5.2
9.7
A symmetrical three-phase; threewire 440 V supply goes to a star-connected load. The impedances
in each branch are ZR = 2 + j3 , ZY = 1– j2 and ZB = 3 + j4 . Find its equivalent delta
connected load.
[AU May/June 2014]
A asymmetrical three-phase, three-wire 400 V supply is connected to a delta-connected load.
Impedances in each branch are ZRY = 10 30° W, ZYB = 10 45º W, and ZBR = 2.5 60º . Find its
equivalent star-connected load.
[AU May/June 2014]
VOLTAGE, CURRENT, AND POWER IN A STAR CONNECTED SYSTEM
9.7.1 Star-Connected System
Figure 9.21 shows a
balanced
three-phase,
Y-connected
system.
The voltage induced in
each winding is called
the phase voltage (VPh). Likewise VRN, VYN , and VBN
represent the rms values of the induced voltages in
each phase. The voltage available between any pair of
terminals is called the line voltage (VL). Likewise VRY,
VYB and VBR are known as line voltages. The double
subscript notation is purposefully used to represent
voltages and currents in polyphase circuits. Thus, VRY
indicates a voltage V between points R and Y, with R
being positive with respect to the point Y during its
Fig. 9.21
positive half cycle.
Similarly, VYB means that Y is positive with respect to the point B during its positive half cycle; it also
means that VRY 5 – VYR.
LO 6
‰ Voltage Relations The phasors corresponding to the phase voltages constituting a three-phase system can be represented by a phasor diagram as shown in Fig. 9.22.
From Fig. 9.22, considering the lines R, Y, and B, the line voltage VRY is equal to the phasor sum of VRN
and VNY which is also equal to the phasor difference of VRN and VYN (VNY 5 – VYN). Hence, VRY is found by
compounding VRN and VYN reversed. To subtract VYN from VRN, we reverse the phasor VYN and find its phasor
sum with VRN as shown in Fig. 9.22. The two phasors, VRN and – VYN, are equal in length and are 60° apart.
|VRN| 5 – |VYN| 5 VPh

VRY = 2VPh cos 60 / 2 = 3 VPh
Similarly, the line voltage VYB is equal to the phasor difference of VYN and VBN, and is equal to 3 VPh
. The line voltage VBR is equal to the phasor difference of VBN and VRN, and is equal to 3 VPh . Hence, in a
balanced star-connected system
(i) Line voltage 5 3 VPh ,
(ii) All line voltages are equal in magnitude and are displaced by 120°, and
(iii) All line voltages are 30° ahead of their respective phase voltages (from Fig. 9.22).
345
Polyphase Circuits
Fig. 9.22
EXAMPLE 9.10
A symmetrical star-connected system is shown in Fig. 9.23 (a). Calculate the
three line voltages, given VRN 5 230 ∠0°. The phase sequence is RYB.
Solution Since the system is a balanced system, all the phase voltages are
equal in magnitude, but displaced by 120° as shown in Fig. 9.23 (b).
 VRN 5 230 ∠0° V
VYN 5 230 ∠–120° V
VBN 5 230 ∠–240° V
Corresponding line voltages
are equal to 3 times the phase
voltages, and are 30° ahead of the
respective phase voltages.
\
Fig. 9.23 (b)
Fig. 9.23 (a)
VRY = 3 × 230 ∠0 + 30° V = 398.37∠30° V
VYB = 3 × 230 ∠−120° + 30° V = 398.37∠− 90° V
VBR = 3 × 230 ∠− 240° + 30° V = 398.37∠− 210° V
Current Relations Figure 9.24 (a) shows a balanced three-phase, wye-connected system indicating
phase currents and line currents. The arrows placed alongside the currents IR, IY, and IB flowing in the three
phases indicate the directions of currents when they are assumed to be positive and not the directions at
346
Circuits and Networks
that particular instant. The phasor diagram for phase currents with respect to their phase voltages is shown
in Fig. 9.24 (b). All the phase currents are displaced by 120° with respect to each other, ‘f’ is the phase
angle between phase voltage and phase current (lagging load is assumed). For a balanced load, all the phase
currents are equal in magnitude. It can be observed from Fig. 9.24 (a) that each line conductor is connected
in series with its individual phase winding Therefore, the current in a line conductor is the same as that in the
phase to which the line conductor is connected.
 IL 5 IPh 5 IR 5 IY 5 IB
It can be observed from Fig. 9.24 (b) that the angle between the line (phase) current and the corresponding
line voltage is (30 1 f)° for a lagging load. Consequently, if the load is leading, then the angle between the
line (phase) current and corresponding line voltage will be (30 – f)°.
Fig. 9.24
EXAMPLE 9.11
In Fig. 9.24 (a), the value of the current in phase R is IR 5 10 ∠20 A. Calculate the values of the three line currents.
Assume an RYB phase sequence.
Solution In a balanced star-connected system IL 5 IPh, and is displaced by 120°. Therefore the three line
currents are
IR 5 10 ∠20° A
IY 5 10 ∠20° – 120° A 5 10 ∠–100° A
IB 5 10 ∠20° – 240° A 5 10 ∠–220° A
Power in the Star-Connected Network The total active power or true power in the three-phase load
is the sum of the powers in the three phases. For a balanced load, the power in each load is the same; hence,
total power 5 3 3 power in each phase
or P 5 3 3 VPh 3 IPh cos f
It is the usual practice to express the three-phase power in terms of line quantities as follows.
VL = 3 VPh , I L = I Ph
or
P = 3 VL I L cos fW
3 VL I L cos f is the active power in the circuit.
Polyphase Circuits
347
Total reactive power is given by
Q = 3 VL I L sin fVAR
Total apparent power or volt-amperes
= 3 VL I L VA
9.7.2 n-Phase Star System
It is to be noted that star and mesh are general terms applicable to any number of phases; but wye and delta
are special cases of star and mesh when the system is a three-phase system. Consider an n-phase balanced
star system with two adjacent phases as shown in Fig. 9.25 (a). Its vector diagram is shown in Fig. 9.25 (b).
Fig. 9.25
The angle of phase difference between adjacent phase voltages is 360°/n. Let EPh be the voltage of each
phase. The line voltage, i.e. the voltage between A and B is equal to EAB 5 EL 5 EAO 1 EOB. The vector
addition is shown in Fig. 9.25 (c). It is evident that the line current and phase current are same.
EAB 5 EAO 1 EOB
Consider the parallelogram OABC.
OB = OC 2 + OA2 + 2×OA×OC × cos u

360° 
= E 2ph + E 2ph + 2 E 2ph cos 180° −


n 
= 2 E 2ph 2 E 2ph cos
= 2 E ph
360
n

180° 
1− cos 2 

 n 


180° 
= 2 E ph 2 sin 2 
 n 
Fig. 9.25 (c)
E L = 2 E ph sin
180
n
348
Circuits and Networks
The above equation is a general equation for line voltage, for example, for a three-phase system, n 5 3;
EL 5 2 Eph sin 60° 5 3 E Ph .
EXAMPLE 9.12
A balanced star-connected load of (4 1 J3) V per phase is connected to a balanced 3-phase 400 V supply.
The phase current is 12 A. Find (a) the total active power, (b) reactive power, and (c) total apparent power.
Solution The voltage given in the data is always the rms value of the line voltage unless otherwise specified.
\
Z Ph = 4 2 + 32 = 5 V
R
4
PF = cos f = Ph = = 0.8
5
Z Ph
sin f = 0.6
(a) Active power
5 3 VLIL cos f W
5 3 3 400 3 12 3 0.8 5 6651 W
(b) Reactive power 5 3 VL IL sin f VAR
5 3 3 400 3 12 3 0.6 5 4988.36 VAR
(c) Apparent power 5 3 VLIL
5
3 3 400 3 12 5 8313.84 VA
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 6
Three non-reactive resistors of 5 V, 10 V and 15 V are star-connected to R, Y and B phase of a 440
V symmetrical system. Determine the current and power in each resistor and the voltage between
star point and neutral; assume the phase sequence as RYB.
rrr 9-6.2 A three-phase, three-wire symmetrical 440 V source is supplying power to an unbalanced, deltaconnected load in which ZRY 5 20 ∠30° V, ZYB 5 20 ∠0° V and ZBR 5 20 ∠– 30° V. If the phase
sequence is RYB, calculate the line currents.
rrr 9-6.3 Three non-inductive resistances of 25 V, 10 V and 15 V are connected in star to a 400 V
symmetrical supply. Calculate the line currents and the voltage across the each load phase.
rrr 9-6.4 Three impedances Z1 5 (10 1 j0) V; Z2 5 (3 1 j4) V and (0 – j10) V are connected in star across
a balanced line voltage of 100 volts. Find the neutral shift voltage between supply and load. Use
Millsman’s theorem.
rrr 9-6.1
Frequently Asked Questions linked to LO 6
rrr 9-6.1
rrr 9-6.2
rrr 9-6.3
A balanced star-connected load having an impedance
15 + j20 per phase is connected to 3f, 440 V, 50 H Z. Find
the line current and power absorbed by the load.
[AU May/June 2014]
A star-connected balanced load draws a current of 35 A per
phase when connected to a 440 V supply. Determine the
apparent power.
[AU May/June 2014]
For the circuit shown in Q.3, calculate the line current, the
power and the power factor. The values of R, L, and C in
each phase are 10 , 1 H and 100 F respectively.
[AU Nov./Dec. 2012]
Fig. Q.3
349
Polyphase Circuits
9.8
VOLTAGE, CURRENT, AND POWER IN A DELTA-CONNECTED SYSTEM
9.8.1 Delta-Connected System
Figure 9.26 shows a balanced three-phase, three-wire, delta-connected system.
LO 7
This arrangement is referred to as mesh connection because it forms a closed
circuit. It is also known as delta connection because the three branches in the
circuit can also be arranged in the shape of delta (D).
From the manner of
interconnection of the three phases in the circuit, it may
appear that the three phases are short-circuited among
themselves. However, this is not the case. Since the system
is balanced, the sum of the three voltages round the closed
mesh is zero; consequently, no current can flow around the
mesh when the terminals are open.
The arrows placed alongside the voltages, VRY, VYB and
Fig. 9.26
VBR, of the three phases indicate that the terminals R, Y and
B are positive with respect to Y, B and R, respectively, during their respective positive half cycles.
‰ Voltage Relations From Fig. 9.27, we notice that only
one phase is connected between any two lines. Hence, the voltage
between any two lines (VL) is equal to the phase voltage (VPh).
 VRY 5 VL 5 VPh
Since the system is balanced, all the phase voltages are equal,
but displaced by 120° from one another as shown in the phasor
diagram in Fig. 9.27. The phase sequence RYB is assumed.
 |VRY| 5 |VYB| 5 |VBR| 5 VL 5 VPh
Fig. 9.27
EXAMPLE 9.13
In Fig. 9.27, the voltage across the terminals R and Y is 400 ∠0 . Calculate the values of the three line voltages.
Assume RYB phase sequence.
Solution In a balanced delta-connected system, |VL| 5 |VPh|, and is displaced by 120°; therefore, the three
line voltages are
VRY 5 400 ∠0° V
VYB 5 400 ∠–120° V
VBR 5 400 ∠–240° V
Current Relations In Fig. 9.28, we notice that since the system is balanced, the three phase currents (IPh), i.e. IR, IY, IB are equal in magnitude but displaced by 120° from one another as shown in Fig.
9.28 (b). I1, I2, and I3 are the line currents (IL), i.e. I1 is the line current in the line 1 connected to the common point of R. Similarly, I2 and I3 are the line currents in lines 2 and 3, connected to common points Y
and B, respectively. Though here all the line currents are directed outwards, at no instant will all the three
350
Circuits and Networks
(a)
(a)
Fig. 9.28
line currents flow in the same direction, either outwards or inwards. Because the three line currents are
displaced 120° from one another, when one is positive, the other two might both be negative, or one positive and one negative. Also it is to be noted that arrows placed alongside phase currents in Fig. 9.28 (a),
indicate the direction of currents when they are assumed to be positive and not their actual direction at a
particular instant. We can easily determine the line currents in Fig. 9.28 (a), I1, I2, and I3 by applying KCL
at the three terminals R, Y and B, respectively. Thus, the current in line 1, I1 5 IR – IB; i.e. the current in
any line is equal to the phasor difference of the currents in the two phases attached to that line. Similarly,
the current in the line 2, I2 5 IY – IR, and the current in the line 3, I3 5 IB – IY.
The phasor addition of these currents is shown in Fig. 9.28 (b). From the figure,
I1 5 IR – IB
I1 = I R2 + I B2 + 2 I R I B cos 60°
I1 5
3 IPh, since IR 5 IB 5 IPh
Similarly, the remaining two line currents, I2 and I3, are also equal to 3 times the phase currents; i.e.
IL 5 3 IPh.
As can be seen from Fig. 9.28 (b), all the line currents are equal in magnitude but displaced by 120° from
one another; and the line currents are 30° behind the respective phase currents.
351
Polyphase Circuits
EXAMPLE 9.14
Three identical loads are connected in delta to a
three-phase supply of 440 ∠0 V as shown in Fig.
9.29 (a). If the phase current IR is 15 ∠0 A, calculate
the three line currents.
All the line currents are equal and 30
behind their respective phase currents, and 3
times their phase values, displaced by 120 from one
another, assuming RYB phase sequence.
Let the line currents in line 1, 2, and 3 be I1, I2 and
I3, respectively.
Solution
I1 5
5
(a)
3 3 IR ∠(f – 30°)
3 3 15 ∠– 30° 5 25.98 ∠– 30° A
I2 5
3 3 15 ∠(– 30 – 120)° 5 25.98 ∠– 150° A
I3 5
3 3 15 ∠(– 30 – 240)° 5 25.98 ∠– 270° A
The phasor diagram is shown in Fig. 9.29 (b).
Power in the Delta-Connected System Obviously, the total power in the delta circuit is the sum of
the powers in the three phases. Since the load is balanced,
the power consumed in each phase is the same. Total
power is equal to three times the power in each phase.
Power per phase 5 VPh IPh cos f
where f is the phase angle between phase voltage and
phase current.
Total power P 5 3 3 VPh IPh cos f
In terms of line quantities,
P5
Since
3 VL IL cos f W
VPh = VL and I Ph =
IL
3
for a balanced system, whether star or delta, the expression
for the total power is the same.
(b)
Fig. 9.29
EXAMPLE 9.15
A balanced delta-connected load of (2 1 j3) V per phase is connected to a balanced three-phase 440 V supply.
The phase current is 10 A. Find the (a) total active power (b) reactive power, and (c) apparent power in the circuit.
Solution
Z Ph =
2
2
(2) + (3) = 3.6 ∠56.3° V
352
Circuits and Networks
cos f =
So,
RPh
2
=
= 0.55
Z Ph 3.6
sin f 5 0.83
IL 5
(a) Active power
(b) Reactive power
3 3 IPh 5
3 3 10 5 17.32 A
5
3 VLIL cos f
5
3 3 440 3 17.32 3 0.55 5 7259.78 W
5
3 VLIL sin f
5
3 3 440 3 17.32 3 0.83 5 10955.67 VAR
(c) Apparent power 5
5
3 VLIL
3 3 440 3 17.32 5 13199.61 VA
9.8.2 n-Phase Mesh System
Figure 9.30 (a) shows part of an n-phase balanced mesh system. Its vector diagram is shown in Fig. 9.30 (b).
Let the current in line BB9 be IL. This is same in all the remaining lines of the n-phase system. IAB, IBC are
the phase currents in AB and BC phases respectively. The vector addition of the line current is shown in Fig.
9.30 (c). It is evident from the Fig. 9.30 (b) that the line and phase voltages are equal.
(b)
(a)
(c)
Fig. 9.30
Polyphase Circuits
353
IBB5 IL 5 IAB 1 ICB
5 IAB – IBC
Consider the parallelogram OABC.

360 
OB = OA2 + OC 2 + 2×OA×OC × cos 180 −


n 
2
2
2
= I Ph
I Ph
2 I Ph
cos
360
n
180 
= 2 I Ph 1 cos 2 
 n 
= 2 I Ph 2 sin 2
180
n
180
n
The above equation is a general equation for the line current in a balanced n-phase mesh system.
I L = 2 I Ph sin
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 7
rrr 9-7.1
rrr 9-7.2
rrr9-7.3
rrr 9-7.4
rrr9-7.5
Three equal resistances connected in star across a three-phase balanced supply consume 1000 W.
If the same three resistors were reconnected in delta across the same supply, determine the power
consumed.
The currents in RY, YB, and BR branches of a mesh connected system with symmetrical voltages are
20 A at 0.7 lagging power factor, 20 A at 0.8 leading power factor, and 10 A at UPF respectively.
Determine the current in each line. Phase sequence is RYB. Draw a phasor diagram.
Three identical impedances 10 ∠30° V in a delta connection, and three identical impedances
5 ∠35° V in a star connection are on the same three-phase, three-wire 173 V system. Find the line
currents and the total power.
Three capacitors, each of 100 mF are connected in delta to a 440 V, three -phase, 50 Hz supply.
What will be the capacitance of each of the three capacitors if the same three capacitors are
connected in star across the same supply to draw the same line current.
Three impedances, ZR 5 (3 1 j2) V; ZY 5 j9 V and ZB 5 3 V are connected in star across a
400 V, 3-wire system. Find the loads on the equivalent delta-connected system phase-sequence RYB.
Frequently Asked Questions linked to LO 7
rrr 9-7.1
rrr 9-7.2
rrr 9-7.3
rrr 9-7.4
rrr 9-7.5
A delta-connected load has (30 + j40) impedance per phase. Determine the phase current if it is
connected to a 415 V, 3-phase, 50 Hz supply.
[AU May/June 2013]
Prove that the total instantaneous power in a balanced three-phase system is constant and is equal
to the average power whether the load is star or delta-connected.
[AU May/June 2013]
A three-phase balanced delta-connected load of 4 + j8 is connected across a 400 V, 3 f balanced
supply. Determine the phase currents and line currents. (Phase sequence is RYB).
[AU May/June 2014]
Write the relation the line and phase value of voltage and current in a balanced delta-connected
system.
[AU May/June 2014]
A 3-phase, 3-wire 120 V RYB system feeds a -connected load whose phase impedance is 30 45º
. Find the phase and line current in this system and draw the phasor diagram. [AU Nov./Dec. 2012]
354
9.9‰
Circuits and Networks
THREE-PHASE BALANCED CIRCUITS
The analysis of three-phase balanced systems is presented in this section. It is no
way different from the analysis of ac systems in general. The relation between
voltages, currents and power in delta-connected and star-connected systems
has already been discussed in the previous sections. We can make use of those
relations and expressions while solving the circuits.
LO 8
9.9.1 Balanced Three-Phase System-Delta Load
Figure 9.31 (a) shows a three-phase, three-wire, balanced system supplying power to a balanced three-phase
delta load. The phase sequence is RYB. We are required to find out the currents in all branches and lines.
Let us assume the line voltage VRY 5 V ∠0° as the reference phasor. Then the three source voltages are
given by
VRY 5 V ∠0° V
VYB 5 V ∠–120° V
VBR 5 V ∠–240° V
These voltages are represented by phasors in Fig. 9.31 (b). Since the load is delta-connected, the line
voltage of the source is equal to the phase voltage of the load. The current in phase RY, IR will lag (lead)
behind (ahead of ) the phase voltage VRY by an angle f as dictated by the nature of the load impedance. The
angle of lag of IY with respect to VYB, as well as the angle of lag of IB with respect to VBR will be f as the load
is balanced. All these quantities are represented in Fig. 9.31 (b).
Fig. 9.31
If the load impedance is Z ∠f, the current flowing in the three load impedances are then
V −0 V
I R = RY
= ∠− f
Z −f
Z
IY =
VYB −120 V
= ∠−120°− f
Z −f
Z
IB =
VBR − 240 V
= ∠ − 240°− f
Z −f
Z
Polyphase Circuits
355
The line currents are 3 times the phase currents and are 30° behind their respective phase currents.
 current in the line 1 is given by
V
∠(−f − 30°), or
Z
Similarly, the current in the line 2
I1 = 3
I R − I B ( phasor difference)
V
∠(−120 − f − 30°),
Z
V
or IY − I R ( phasor difference) = 3 ∠ (−f −150°) , and
Z
V
I 3 = 3 ∠(−240 − f − 30°), or I B − IY ( phasor difference)
Z
V
= 3 − (−270 − f)
Z
To draw all these quantities vectorially, VRY 5 V∠0° is taken as the reference vector.
I2 = 3
EXAMPLE 9.16
A three-phase, balanced delta-connected load of (4 1 j8) V is connected across a 400 V, 3 – f balanced
supply. Determine the phase currents and line currents. Assume the phase sequence to be RYB. Also calculate
the power drawn by the load.
Solution Refering to Fig. 9.31 (a), taking the line voltage VRY 5 V∠0° as reference VRY 5 400 ∠0° V;
VYB 5 400 ∠–120° V, VBR 5 400 ∠–240° V
Impedance per phase 5 (4 1 j8) V 5 8.94 ∠63.4° V
Phase currents are: I R =
400 ∠0°
= 44.74 ∠− 63.4° A
8.94 ∠63.4°
IY =
400 −120
= 44.74 ∠−183.4° A
8.94 − 63.4
400 − 240
= 44.74 ∠− 303.4° A
8.94 − 63.4
The three line currents are
IB =
I1 5 IR – IB 5 (44.74 ∠–63.4° –44.74 ∠–303.4°)
5 (20.03 – j40) – (24.62 1 j37.35) 5 (–4.59 – j77.35) A
5 77.49 ∠266.6° A
or the line current I1 is equal to the
I1 5
or
3 times the phase current and 30° behind its respective phase current
3 3 44.74 ∠–63.4° –30° 5 77.49 ∠–93.4°
5 77.49 ∠266.6° A
Similarly, I2 5 IY – IR
5
3 3 44.74 ∠–183.4° –30° 5 77.49 –213.4° A
356
Circuits and Networks
5 77.49 ∠146.6° A
I3 5 IB – IY
5
3 3 44.74 ∠– 303.4° –30° 5 77.49 –333.4° A
5 77.49 ∠26.6° A
Power drawn by the load is P 5 3VPh IPh cos f
or
3 3 VL 3 IL cos 63.4° 5 24.039 kW
9.9.2 Balanced Three Phase System-Star Connected Load
Figure 9.32 (a) shows a three-phase, three-wire system supplying power to a balanced three phase star
connected load. The phase sequence RYB is assumed.
In star connection, whatever current is flowing in the phase is also flowing in the line. The three line
(phase) currents are IR, IY, and IB.
Fig. 9.32 (a)
VRN, VYN, and VBN represent three phase voltages of the network, i.e. the voltage between any line and
neutral. Let us assume the voltage VRN 5 V∠0° as the reference phasor. Consequently, the phase voltage
VRN 5 V ∠0°
VYN 5 V ∠–120°
VBN 5 V ∠–240°
Hence I R =
VRN
V −0
V
=
= ∠−f
Z ∠f Z − f
Z
IY =
VYN
V − 120 V
=
= ∠ −120° − f
Z −f
Z −f
Z
IB =
VBN
V − 240 V
=
= ∠ − 240° − f
Z −f
Z −f
Z
Polyphase Circuits
357
As seen from the above expressions, the currents, IR,
IY, and IB, are equal in magnitude and have a 120° phase
difference. The disposition of these vectors is shown in
Fig. 9.32 (b). Sometimes, a fourth wire, called neutral
wire is run from the neutral point, if the source is also
star-connected. This gives three-phase, four-wire starconnected system. However, if the three line currents are
balanced, the current in the fourth wire is zero; removing
this connecting wire between the source neutral and load
neutral is, therefore, not going to make any change in the
condition of the system. The availability of the neutral
wire makes it possible to use all the three phase voltages,
as well as the three line voltages. Usually, the neutral is
grounded for safety and for the design of insulation.
It makes no difference to the current flowing in the
load phases, as well as to the line currents, whether the
Fig. 9.32 (b)
sources have been connected in star or in delta, provided
the voltage across each phase of the delta connected source is 3 times the voltage across each phase of the
star-connected source.
EXAMPLE 9.17
A balanced star-connected load having an impedance (15 1 j20) V per phase is connected to a three-phase,
440 V; 50 Hz supply. Find the line currents and the power absorbed by the load. Assume RYB phase sequence.
Solution Referring to Fig. 9.32 (a), taking VRN as the reference voltage, we have
440 ∠0°
VRN =
= 254 ∠0°
3
VYN 5 254 ∠–120°
VBN 5 254 ∠–240°
Impedance per phase, ZPh 5 15 1 j20 5 25 ∠53.13° V
VRN
254 − 0
=
= 10.16 ∠−53.13° A
Z Ph
25 − 53.13
V
254 −120
IY = YN =
= 10.16 ∠−173.13° A
Z Ph
25 − 53.13
V
254 − 240
I B = BN =
= 10.16 ∠−293.13° A
Z Ph
25 − 53.13
The three phase currents are equal in magnitude and displaced by 120° from one another. Since the load
is star-connected, these currents also represents line currents.
The power absorbed by the load (P)
The phase currents are
or
IR =
5
3 3 VPh 3 IPh cos f
5
3 3 VL 3 IL cos f
5
3 3 440 3 10.16 3 cos 53.13° 5 4645.78 W
358
9.10
Circuits and Networks
THREE-PHASE-UNBALANCED CIRCUITS
LO 8
9.10.1 Types of Unbalanced Loads
An unbalance exists in a circuit when the impedances in one or more phases differ from the impedances of
the other phases. In such a case, line or phase currents are different and are displaced from one another by
unequal angles. So far, we have considered balanced loads connected to balanced systems. It is enough to solve
problems, considering one phase only on balanced loads; the conditions on other two phases being similar.
Problems on unbalanced three-phase loads are difficult to handle because conditions in the three phases are
different. However, the source voltages are assumed to be balanced. If the system is a three-wire system, the
currents f lowing towards the load in the three lines must add to zero at any given instant. If the system is a
four-wire system, the sum of the three outgoing line currents is equal to the return current in the neutral wire.
We will now consider different methods to handle unbalanced star-connected and delta-connected loads. In
practice, we may come across the following unbalanced loads:
1. Unbalanced delta-connected load,
2. Unbalanced three-wire star-connected load, and
3. Unbalanced four-wire star-connected load.
Unbalanced Delta-connected Load Figure 9.33
shows an unbalanced delta-load connected to a balanced
three-phase supply.
The unbalanced delta-connected load supplied from
a balanced three-phase supply does not present any new
problems because the voltage across the load phase is
fixed. It is independent of the nature of the load and is
equal to the line voltage of the supply. The current in
each load phase is equal to the line voltage divided by
the impedance of that phase. The line current will be the
phasor difference of the corresponding phase currents,
taking VRY as the reference phasor.
Assuming RYB phase sequence, we have
VRY 5 V ∠0° V, VYB 5 V ∠–120° V, VBR 5 V ∠–240° V
Phase currents are
V
V ∠0°
V
I R = RY =
=
∠− f1 A
Z1 ∠f Z1 ∠ f1
Z1
IY =
VYB
V ∠−120°
V
=
=
∠ −120° − f 2 A
Z2 ∠ f2
Z2 ∠ f2
Z2
IB =
VBR
V ∠− 240° V
=
=
∠− 240° − f3 A
Z3 ∠ f3
Z3 ∠ f3
Z3
The three line currents are
I1 5 IR – IB phasor difference
I2 5 IY – IR phasor difference
I3 5 IB – IY phasor difference
Fig. 9.33
Polyphase Circuits
359
EXAMPLE 9.18
Three impedances Z1 5 20 ∠30° V, Z2 5 40 ∠60° V and Z3 5 10 ∠–90° V are delta-connected to a 400 V,
3 – f system as shown in Fig. 9.34. Determine the (a) phase currents, (b) line currents, and (c) total power
consumed by the load.
Fig. 9.34
Solution The three phase currents are IR, IY, and IB, and the three line currents are I1, I2 and I3. Taking
VRY 5 V ∠0° V as reference phasor, and assuming RYB phase sequence, we have
VRY 5 400 ∠0° V, VYB 5 400 ∠–120° V,
VBR 5 400 ∠–240° V
Z1 5 20 ∠30° V 5 (17.32 1 j10) V;
Z2 5 40 ∠60° V 5 (20 1 j34.64) V;
(a)
Z3 5 10 ∠–90° V 5 (0 – j10) V
V
400∠0°
I R = RY =
A = 20 ∠− 30° A
Z1 ∠f1 20∠30°
5 (17.32 – j10) A
IY =
VYB
400 ∠−120°
=
A = 10 ∠−180° A
Z 2∠f 2
40 ∠60°
5 (–10 1 j0) A
IB =
VBR
400 ∠− 240°
=
A = 40 ∠−150° A
Z3 ∠f3
10 ∠− 90°
5 (–34.64 – j20) A
(b) Now, the three line currents are
I1 5 IR – IB 5 [(17.32 – j10) – (–34.64 – j20)]
5 (51.96 1 j10) A 5 52.91 ∠10.89° A
I2 5 IY – IR 5 [(–10 1 j0) – (17.32 – j10)]
5 (–27.32 1 j10) A 5 29.09 ∠159.89° A
I3 5 IB – IY 5 [(–34.64 – j20) – (–10 1 j0)]
5 (–24.64 – j20) A 5 31.73 ∠–140.94° A
360
Circuits and Networks
(c) To calculate the total power, first the powers in the individual phases are to be calculated, then
they are added up to get the total power in the unbalanced load.
Power in R phase 5 I 2 R 3 RR 5 (20)2 3 17.32 5 6928 W
Power in Y phase 5 I 2 Y 3 RY 5 (10)2 3 20 5 2000 W
Power in B phase 5 I 2 B 3 RB 5 (40)2 3 0 5 0
 total power in the load 5 6928 1 2000 5 8928 W
Unbalanced Four-Wire Star-Connected Load Figure 9.35 shows an unbalanced star load connected to a balanced 3-phase, 4-wire supply.
Fig. 9.35
The star point, NL, of the load is connected to the star point, NS of the supply. It is the simplest case of
an unbalanced load because of the presence of the neutral wire; the star points of the supply NS (generator)
and the load NL are at the same potential. It means that the voltage across each load impedance is equal to
the phase voltage of the supply (generator), i.e. the voltages across the three load impedances are equalised
even though load impedances are unequal. However, the current in each phase (or line) will be different.
Obviously, the vector sum of the currents in the three lines is not zero, but is equal to neutral current. Phase
currents can be calculated in similar way as that followed in an unbalanced delta-connected load.
Taking the phase voltage VRN 5 V∠0° V as reference, and assuming RYB phase sequences, we have the
three phase voltages as follows:
VRN 5 V ∠0° V, VYN 5 V ∠–120° V, VBN 5 V ∠–240° V
The phase currents are
IR =
VRN
V ∠0°
V
=
A=
∠− f1 A
Z1
Z1 ∠f1
Z1
IY =
VYN V ∠120°
V
=
∠−120° − f 2 A
A=
Z2
Z 2 ∠f 2
Z2
IB =
VBN V ∠240°
V
=
∠− 240° − f3 A
A=
Z3
Z3∠f3
Z3
Polyphase Circuits
361
Incidentally, IR, IY and IB are also the line currents; the current in the neutral wire is the vector sum of the
three line currents.
EXAMPLE 9.19
An unbalanced four-wire, star-connected load has a balanced voltage of 400 V, the loads are
Z1 5 (4 1 j8) V; Z2 5 (3 1 j4) V; Z3 5 (15 1 j20) V
Calculate the (a) line currents, (b) current in the neutral wire, and (c) the total power.
Solution
Z1 5 (4 1 j8) V; Z2 5 (3 1 j4) V; Z3 5 (15 1 j20) V
Z1 5 8.94 ∠63.40° V; Z2 5 5 ∠53.1° V; Z3 5 25 ∠53.13° V
Let us assume RYB phase sequence.
The phase voltage VRN =
400
= 230.94V
3
Taking VRN as the reference phasor, we have
VRN 5 230.94 ∠0° V, VYN 5 230.94 ∠–120° V
VBN 5 230.94 ∠–240° V
The three line currents are
(a)
IR =
VRN
230.94 ∠0°
=
A = 25.83 ∠− 63.4° A
8.94 ∠63.4°
Z1
IY =
VYN
230.94∠−120°
=
A = 46.188 ∠−173.1° A
5∠53.1
Z2
IB =
VBN
230.94∠−120°
=
A = 9.23 ∠− 293.13° A
25∠ 53.13
Z3
(b) To find the neutral current, we must add the three line currents. The neutral current must then be
equal and opposite to this sum.
Thus, IN 5 – (IR 1 IY 1 IB)
5 – (25.83 ∠–63.4° 1 46.188 ∠–173.1° 1 9.23 ∠–293.13°) A
IN 5 – [(11.56 – j23.09) 1 (–45.85 – j5.54) 1 (3.62 1 j8.48)] A
IN 5 – [(–30.67 – j20.15)] A 5 (30.67 1 j20.15) A
IN 5 36.69 ∠33.30° A
Its phase with respect to VRN is 33.3°, the disposition of all the currents is shown in Fig. 9.36.
(c) Power in R phase 5 I 2R 3 RR 5 (25.83)2 3 4 5 2668.75 W
Power in Y phase 5 I 2Y 3 RY 5 (46.18)2 3 3 5 6397.77 W
362
Circuits and Networks
Fig. 9.36
Power in B phase 5 I 2B 3 RB 5 (9.23)2 3 15 5 1277.89 W
Total power absorbed by the load
5 2668.75 1 6397.77 1 1277.89 5 10344.41 W
Unbalanced Three-Wire Star-Connected Load In a three-phase, four-wire system if the connection
between supply neutral and load neutral is broken, it would result in an unbalanced three-wire star-load. This
type of load is rarely found in practice, because all the three-wire star loads are balanced. Such as system is
shown in Fig. 9.37. Note that the supply star point (NS) is isolated from the load star point (NL). The potential
of the load star point is different from that of the supply star point. The result is that the load phase voltages
is not equal to the supply phase voltage; and they are not only unequal in magnitude, but also subtend angles
other than 120° with one another. The magnitude of each phase voltage depends upon the individual phase
loads. The potential of the load neutral point changes according to changes in the impedances of the phases,
that is why sometimes the load neutral is also called a floating neutral point. All star-connected, unbalanced
loads supplied from polyphase systems without a neutral wire have f loating neutral point. The phasor sum of
the three unbalanced line currents is zero. The phase voltage of the load is not 1 / 3 of the line voltage. The
Fig. 9.37
Polyphase Circuits
363
unbalanced three-wire star load is difficult to deal with. It is because load phase voltages cannot be determined
directly from the given supply line voltages. There are many methods to solve such unbalanced Y-connected
loads. Two frequently used methods are presented here. They are
1. Star-delta conversion method, and
2. The application of Millman’s theorem
9.10.2 Star-Delta Method of Solving Unbalanced Load
Figure 9.38 (a) shows an unbalanced wye-connected load. It has already been shown in Section 9.6 that a
three-phase star-connected load can be replaced by an equivalent delta-connected load. Thus, the star load
of Fig. 9.38 (a) can be replaced by equivalent delta as shown in Fig. 9.38 (b), where the impedances in each
phase is given by
Z Z + ZY Z B + Z B Z R
Z RY = R Y
ZB
Z R ZY + ZY Z B + Z B Z R
ZYB =
ZR
Z R ZY + ZY Z B + Z B Z R
Z BR =
ZY
Fig. 9.38
The problem is then solved as an unbalanced delta-connected system. The line currents so calculated are
equal in magnitude and phase to those taken by the original unbalanced wye (Y ) connected load.
EXAMPLE 9.20
A 400 V, three-phase supply feeds an unbalanced three-wire, star-connected load. The branch impedances of
the load are ZR 5 (4 1 j8) V; Zy 5 (3 1 j4) V and ZB 5 (15 1 j20) V. Find the line currents and voltage
across each phase impedance. Assume RYB phase sequence.
Solution The unbalanced star load and its equivalent delta (D) is shown in Fig. 9.39 (a) and (b) respectively.
ZR 5 (4 1 j8) V 5 8.944 ∠63.4° V
ZY 5 (3 1 j4) V 5 5 ∠53.1° V
ZB 5 (15 1 j20) V 5 25 ∠53.1° V
364
Circuits and Networks
Fig. 9.39
Using the expression in Section 9.10.2, we can calculate ZRY, ZYB and ZBR
ZRZY 1 ZYZB 1 ZBZR
5 (8.94 ∠63.4°) (5 ∠53.1°) 1 (5 ∠53.1°) (25 ∠53.1°)
1 (25 ∠53.1°) (8.94 ∠63.4°)
5 391.80 ∠113.23°
Z RY =
Z R ZY + ZY Z B + Z B Z R 391.80 ∠113.23°
=
= 15.67 ∠60.13°
ZB
25 ∠53.1°
ZYB =
Z R ZY + ZY Z B + Z B Z R 391.80 ∠113.23°
= 43.83 ∠49.83°
=
ZR
8.94 ∠63.4°
Z BR =
Z R ZY + ZY Z B + Z B Z R 391.80 ∠113.23°
= 78.36 ∠60.13°
=
ZY
5 ∠53.1°
Taking VRY as reference, VRY 5 400 ∠0
VYB 5 400 ∠–120°; VBR 5 400 ∠–240°
IR =
VRY
400∠0°
=
= 25.52 ∠− 60.13°
Z RY
15.67∠60.13°
IY =
VYB
400 ∠−120°
=
= 9.12 ∠−169.83°
ZYB
43.83 ∠49.83°
IB =
VBR
400 ∠− 240°
=
= 5.10 ∠− 300.13°
Z BR 78.36 ∠60.13°
The various line currents in the delta load are
I1 5 IR – IB 5 25.52 ∠–60.13° – 5.1 ∠–300.13°
5 28.41 ∠–69.07° A
Polyphase Circuits
365
I2 5 IY – IR 5 9.12 ∠–169.83° – 5.52 ∠–60.13°
5 29.85 ∠136.58° A
I3 5 IB – IY 5 5.1 ∠–300.13° – 9.12 ∠–169.83°
5 13 ∠27.60° A
These line currents are also equal to the line (phase) currents of the original star-connected load. The voltage
drop across each star-connected load will be as follows.
Voltage drop across ZR 5 I1ZR
5 (28.41 ∠–69.070°) (8.94 ∠63.4°) 5 253.89 ∠–5.67° V
Voltage drop across ZY 5 I2ZY
5 (29.85 ∠136.58°) (5 ∠53.1°) 5 149.2 ∠189.68° V
Voltage drop across ZB 5 I3ZB
5 (13 ∠27.60°) (25 ∠53.1°) 5 325 ∠80.70° V
9.10.3 Millman’s Method of Solving Unbalanced Load
One method of solving an unbalanced three-wire star-connected load by star-delta conversion is described in
Section 9.10.2. But this method is laborious and involves lengthy calculations. By using Millman’s theorem,
we can solve this type of problems in a much easier way. Consider an unbalanced wye (Y ) load connected
to a balanced three-phase supply as shown in Fig. 9.40 (a). VRO, VYO and VBO are the phase voltages of the
supply. They are equal in magnitude, but displaced by 120° from one another. VRO9, VYO9 and VBO9 are the
load phase voltages; they are unequal in magnitude as well as differ in phase by unequal angles. ZR, ZY and
ZB are the impedances of the branches of the unbalanced wye (Y ) connected load. Figure 9.40 (b) shows the
triangular phasor diagram of the complete system. Distances RY, YB, and BR represent the line voltages of
the supply as well as load. They are equal in magnitude, but displaced by 120°. Here, O is the star-point of
the supply and is located at the centre of the equilateral triangle RYB. O9 is the load star point. The star point
of the supply which is at the zero potential is different from that of the star point at the load, due to the load
being unbalanced. O9 has some potential with respect to O and is shifted away from the centre of the triangle.
Distance O9O represents the voltage of the load star point with respect to the star point of the supply Vo9o.
Fig. 9.40 (a)
366
Circuits and Networks
Vo9o is calculated using Millman’s theorem. If Vo9o is known, the load phase voltages and corresponding
currents in the unbalanced wye load can be easily determined.
According to Millman’s theorem, Vo9o is given by
Vo′o =
VRo YR + VYo YY + VBo YB
YR + YY + YB
where the parameters YR, YY, and YB are the admittances of the branches
of the unbalanced wye connected load. From Fig. 9.40 (a), we can write
the equation
VRo 5 VRo9 1 Vo9o
or the load phase voltage
Fig. 9.40 (b)
VRo9 5 VRo – Vo9o
Similarly, VYo9 5 VYo – Vo9o and VBo9 5 VB o– Vo9o can be calculated. The line currents in the load are
IR =
VRo′
= (VRo −Vo′o )YR
ZR
IY =
VYo′
= (VYo −Vo′o )YY
ZY
IB =
VBo′
= (VBo −Vo′o )YB
ZB
The unbalanced three-wire star-connected loads can also be determined by using Kirchhoff’s laws, and
Maxwell's mesh or loop equation. In general, any method which gives quick results in a particular case should
be used.
EXAMPLE 9.21
To illustrate the application of Millman's method to unbalanced loads, let us take the problem in example
given in Section 9.10.2.
Solution The circuit diagram is shown in Fig. 9.41.
Fig. 9.41
Polyphase Circuits
367
Taking VRY as reference line voltage 5 400 ∠0°, phase voltages lag 30° behind their respective line
voltages. Therefore, the three phase voltages are
400
VRo =
∠− 30° V
3
VYo =
VBo =
400
3
400
3
∠−150° V
∠− 270°V
The admittances of the branches of the wye load are
YR =
1
1
=
= 0.11 ∠− 63.40° �
Z R 8.94 ∠63.4°
YY =
1
1
=
= 0.2 ∠− 53.1° �
ZY
5 ∠53.1°
YB =
1
1
=
= 0.04 ∠− 53.1° �
Z B 25 ∠53.1°
VRoYR 1 VYoYY 1 VBoYB 5 (230.94 ∠–30°) (0.11 ∠–63.40°)
1 (230.94 ∠–150°) (0.2 ∠–53.1°)
1 (230.94 ∠–270°) (0.04 ∠–53.1°) 5 36.68 ∠182.66°
YR 1 YY 1 YB 5 0.11 ∠–63.4° 1 0.2 ∠–53.1° 1 0.04 ∠–53.1°
5 0.35 ∠–56.2°
Substituting the above values in the Millman's theorem, we have
Vo′o =
=
VRo YR +VYo YY +VBo YB
YR + YY + YB
36.68 ∠182.66°
= 104.8 ∠238.86°
0.35 ∠− 56.2°
The three load phase voltages are
VRo9 5 VRo – Vo9o
5 230.94 ∠–30° –104.8 ∠238.86° 5 253.89 ∠–5.67° V
VYo9 5 VYo – Vo9o
5 230.94 ∠–150° –104.8 ∠238.86° 5 149.2 ∠189.68° V
VBo9 5 VBo – Vo9o
5 230.94 ∠–270° –104.8 ∠238.86° 5 325 ∠80.7° V
368
Circuits and Networks
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 8
balanced three-phase, star-connected voltage source has VRN 5230 ∠60° V Vrms with RYB
phase sequence, and it supplies a balanced delta-connected three-phase load. The total power
drawn by the load is 15 kW at 0.8 lagging power factor. Find the line currents, load and phase
currents.
rrr 9-8.2 A 400 V, three-phase supply feeds an unbalanced three-wire, star-connected load, consisting of
impedances ZR 5 7 ∠10° V, ZY 5 8 ∠30° V and ZB 5 8 ∠50° V. The phase sequence is RYB.
Determine the line currents and total power taken by the load.
rrr 9-8.3 The circuit shown in Fig. Q.3 is supplied from a 3-f balanced supply. Use PSpice to calculate rms
magnitudes and phase angles of currents: Ia, Ib, Ic, and IN.
rrr 9-8-1 A
Fig. Q.3
A balanced star-connected load is supplied from a symmetrical 3-phase, 440 V; 50 Hz supply. The
current in phase is 20 A and lags its respective phase voltage by an angle 40°. Calculate (a) load
parameters, (b) total power, and (c) readings of two wattmeters in load current to measure total power.
rrr 9-8.5 Consider the unbalanced D – D circuit shown in Fig. Q.5. Use PSpice to find generator currents,
the line currents and phase currents.
rrr 9-8.4
Fig. Q.5
rrr 9-8.6
For the unbalanced circuit shown in Fig. Q.6,
calculate
(a) the line currents,
(b) the real power absorbed by the load, and
(c) the total complex power supplied by the
source. Use PSpice.
Fig. Q.6
369
Polyphase Circuits
Frequently Asked Questions linked to LO8
rrr 9-8.1
A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW
when the line voltage is 220 V and the line current is 18.2 A. Determine the power factor of the
motor.
[AU Nov./Dec. 2012]
rrr 9-8.2
In a three-phase balanced delta system the voltage across R and Y is 400 0ºV. What will be the
voltage across Y and B? Assume RYB phase sequence.
[AU April/May 2011]
rrr 9-8.3
A balanced -connected load has one phase current IBC = 2 –90º A. Find the other phase current
and the three line currents if the system is an ABC system. If the line voltage is 100 V, what is the
load impedance?
[AU April/May 2011]
rrr 9-8.4
The power consumed in a three-phase, balanced star-connected load is 2 kW at a power factor of
0.8 lagging. The supply voltage is 400 V, 50 Hz. Calculate the resistance and reactance of each
phase.
[AU April/May 2011]
rrr 9-8.5
Give examples for balanced networks. Why are they called so?
[PTU 2011-12]
rrr 9-8.6
An unbalanced star-connected load has balanced voltage of 100 V and RBY phase sequence.
Calculate the line currents and the neutral current.
[AU May/June 2013]
Take: ZA = 15 , ZB = (10 + j5) , ZC = (6 – j8) .
rrr 9-8.7
Distinguish between unbalanced source and unbalanced load.
[AU May/June 2013]
rrr 9-8.8
A symmetrical 3-phase, 100 V, 3-wire supply feeds an unbalanced star-connected load with
impedances of the load as ZR = 5 0º , ZY =2 90º , and ZB = 4 –90º . Find the line currents,
voltage across the impedances and draw the phasor diagram. Also calculate the power consumed
by the load.
[AU May/June 2014]
rrr 9-8.9
A three-phase four-wire 120 V ABC system feeds an unbalanced Y-connected load with ZA = 5 0º
, ZB = 10 30º , and ZC = 20 60º , Obtain the four line currents.
[AU Nov./Dec. 2012]
rrr9-8.10 Three impedances Z1 = (17.32 + j10), Z2 = (20 + j34.64), and Z3 = (0 –j10) ohms are deltaconnected to a 400 V, three-phase system. Determine the phase currents, line current, and total
power consumed by the load.
[AU Nov./Dec. 2012]
9.11
POWER MEASUREMENT IN THREE-PHASE CIRCUITS
9.11.1 Power Measurement in a Single-Phase Circuit by Wattmeter
Wattmeters are generally used to measure power in the circuits. A wattmeter
principally consists of two coils, one coil is called the current coil, and
the other the pressure or voltage coil. A diagramatric representation of a
wattmeter connected to measure power in a single-phase circuit is shown
in Fig. 9.42.
Fig. 9.42
LO 9
370
Circuits and Networks
The coil represented with less number of turns between M and L is the current coil, which carries the
current in the load and has very low impedance. The coil with more number of turns between the common
terminal (comn) and V is the pressure coil, which is connected across the load and has high impedance.
The load voltage is impressed across the pressure coil. The terminal M denotes the mains side, L denotes
load side, common denotes the common point of current coil and pressure coil, and V denotes the second
terminal of the pressure coil, usually selected as per the range of the load voltage in the circuit. From the
figure, it is clear that a wattmeter has four terminals, two for current coil and two for potential coil. When
the current f low through the two coils, they set up magnetic fields in space. An electromagnetic torque
is produced by the interaction of the two magnetic fields. Under the influence of the torque, one of the
coils (which is movable) moves on a calibrated scale against the action of a spring. The instantaneous
torque produced by electromagnetic action is proportional to the product of the instantaneous values of
the currents in the two coils. The small current in the pressure coil is equal to the input voltage divided
by the impedance of the pressure coil. The inertia of the moving system does not permit it to follow the
instantaneous f luctuations in torque. The wattmeter def lection is therefore, proportional to the average
power (VI cos f) delivered to the circuit. Sometimes, a wattmeter connected in the circuit to measure
power gives downscale reading or backward
def lection. This is due to improper connection
of the current coil and pressure coil.
To obtain up scale reading, the terminal marked
as ‘Comn’ of the pressure coil is connected to one
of the terminals of the current coil as shown in
Fig. 9.43. Note that the connection between the
current coil terminal and pressure coil terminal
is not inherent, but has to be made externally.
Even with proper connections, sometimes the
wattmeter will give downscale reading whenever
the phase angle between the voltage across the
pressure coil and the current through the current
coil is more than 90°. In such a case, connection
of either the current coil or the pressure coil must
be reversed.
Fig. 9.43
9.11.2 Power in Three-Phase Circuits
Measurement of power by a wattmeter in a single-phase circuit can be extended to measure power in a threephase circuit. From Section 9.11.1, it is clear that we require three wattmeters, one in each phase to measure
the power consumed in a three-phase system. Obviously, the total power is the algebraic sum of the readings
of the three wattmeters. In this way we can measure power in balanced and unbalanced loads. In a balanced
case it would be necessary to measure power only in one phase and the reading is multiplied by three to get
the total power in all the three phases. This is true in principle, but presents a few difficulties in practice. To
verify this fact let us examine the circuit diagram in Figs 9.44 (a) and (b).
Observation of Figs 9.44 (a) and (b) reveals that for a star-connected load, the neutral must be available for
connecting the pressure coil terminals. The current coils must be inserted in each phase for a delta-connected
load. Such connections sometimes may not be practicable, because the neutral terminal is not available all the
time in a star-connected load, and the phases of the delta-connected load are not accessible for connecting the
current coils of the wattmeter. In most of the commercially available practical three-phase loads, only three
line terminals are available. We, therefore, require a method where we can measure power in the three-phases
Polyphase Circuits
371
Fig. 9.44 (a)
Fig. 9.44 (b)
with an access to the three lines connecting the source to the load. Two such methods are discussed here.
9.11.3 Three-Wattmeter and Two-Wattmeter Methods
In this method, the three wattmeters are connected in the three lines as shown in Fig. 9.45, i.e. the current
coils of the three wattmeters are introduced in the three lines, and one terminal of each potential coil is
connected to one terminal of the corresponding current coil, the other three being connected to some common
point which forms an effective neutral n.
The load may be either star-connected or delta-connected. Let us assume a star-connected load, and let the
neutral of this load be denoted by N. Now the reading on the wattmeter WR will correspond to the average value
of the product of the instantaneous value of the current IR f lowing in line 1, with the voltage drop VRn, where
VRn is the voltage between points R and n. This can be written as VRn 5 VRN 1 VNn, where VRN is the load phase
voltage and VNn is the voltage between load neutral, N, and the common point, n. Similarly, VYn 5 VYN 1 VNn,
372
Circuits and Networks
Fig. 9.45
and VBn 5 VBN 1 VNn. Therefore, the average power, WR indicated by the wattmeter is given by
WR =
1
T
T
∫ VRn I R dt
o
where T is the time period of the voltage wave
WR =
1
T
Similarly,
T
∫ (VRN +VNn ) I R dt
o
WY =
=
WB =
and
=
1
T
1
T
1
T
1
T
T
∫ VYn IY dt
o
T
∫ (VYN +VNn ) IY dt
o
T
∫ VBn I B dt
o
T
∫ (VBN +VNn ) I B dt
o
Total average power 5 WR 1 WY 1 WB
T
T
1
1
= ∫ (VRN I R + VYN IY + VBN I B ) dt + ∫ VNn ( I R + IY + I B ) dt
T o
T o
Since the system in the problem is a three-wire system, the sum of the three currents IR, IY, and IB at any
given instant is zero. Hence, the power read by the three wattmeters is given by
WR + WY + WB =
1
T
T
∫ (VRN I R +VYN IY +VBN I B ) dt
o
If the system has a fourth wire, i.e. if the neutral wire is available, then the common point, n is to be
connected to the system neutral, N. In that case, VNn would be zero, and the above equation for power would
Polyphase Circuits
373
still be valid. In other words, whatever be the value of VNn, the algebraic sum of the three currents IR, IY, and
IB is zero. Hence, the term VNn (IR 1 IY 1 IB) would be zero. Keeping this advantage in mind, suppose the
common point, n, in Fig. 9.45 is connected to line B. In such case, VNn 5 VNB; then the voltage across the
potential coil of wattmeter WB will be zero and this wattmeter will read zero. Hence, this can be removed
from the circuit. The total power is read by the remaining two wattmeters, WR and WY.
 total power 5 WR 1 WY
Let us verify this fact from Fig. 9.46.
Fig. 9.46
The average power indicated by the wattmeter WR is
WR =
and that by WY =
Also
T
1
T
∫
1
T
∫
VRB I R dt
o
T
VYB IY dt
o
VRB 5 VRN 1 VNB
VYB 5 VYN 1 VNB
WR + WY =
=
=
We know that
T
1
T
∫ (VRB ⋅ I R +VYB IY ) dt
1
T
∫ {(VRN +VNB ) I R + (VYN +VNB ) IY } dt
1
T
∫ {(VRN I R +VYN IY ) + ( I R + IY )VNB } dt
o
T
o
T
o
IR 1 IY 1 IB 5 0
IR 1 IY 5 – IB
374
Circuits and Networks
Substituting this value in the above equation, we get
WR + WY =
1
T
T
∫ {(VRN I R +VYN IY ) + (−I B )VNB } dt
o
VNB 5 –VBN
T
1
{(VRN I R +VYN ⋅ IY +VBN ⋅ I B )} dt
T ∫o
which indicates the total power in the load.
From the above discussion, it is clear that the power in a three-phase load, whether balanced or unbalanced,
star-connected or delta-connected, three-wire or four-wire, can be measured with only two wattmeters as
shown in Fig. 9.46. In fact, the two wattmeter method of measuring power in three-phase loads has become a
universal method. If neutral wire is available in this method it should not carry any current, or the neutral of
the load should be isolated from the neutral of the source.
The current f lowing through the current coil of each wattmeter is the line current, and the voltage across
the pressure coil is the line voltage. In case the phase angle between line voltage and current is greater
than 90°, the corresponding wattmeter would indicate downscale reading. To obtain upscale reading, the
connections of either the current coil, or the pressure coil has to be interchanged. Reading obtained after
reversal of coil connection should be taken as negative. Then, the algebraic sum of the two wattmeter readings
gives the total power.
WR + WY =
9.11.4 Power Factor by Two-Wattmeter Method
When we talk about the power factor in three-phase circuits, it applies only to balanced circuits, since the
power factor in a balanced load is the power factor of any phase. We cannot strictly define the power factor
in three-phase unbalanced circuits, as every phase has a separate power factor. The two-wattmeter method,
when applied to measure power in a three-phase balanced circuits, provides information that help us to
calculate the power factor of the load.
Figure 9.47 shows the vector diagram of the circuit shown in Fig. 9.46. Since the load is assumed to be
balanced, we can take Z1 ∠f1 5 Z2 ∠f2 5 Z3 ∠f3 5 Z ∠f for the star-connected load. Assuming RYB phase
sequence, the three rms load phase voltages are VRN, VYN and VBN ? IR, IY and IB are the rms line (phase) currents.
These currents will lag behind their respective phase voltages by an angle f. (An inductive load is considered).
Now consider the readings of the two wattmeters in Fig. 9.46.
WR measures the product of effective value of the current through
its current coil IR, effective value of the voltage across its pressure
coil VRB and the cosine of the angle between the phasors IR and
VRB. The voltage across the pressure coil of WR is given as follows.
VRB 5 VRN – VBN phasor difference
It is clear from the phasor diagram that the phase angle
between
VRB and IR is (30° – f)

Fig. 9.47
WR 5 VRB IR cos (30° – f)
Similarly, WY measures the product of effective value of
the current through its current coil IY, the effective value of the
voltage across its pressure coil, VYB and the cosine of the angle
between the phasors VYB and IY.
375
Polyphase Circuits
VYB 5 VYN – VBN
From Fig. 9.47, it is clear that the phase angle between VYB and IY is (30° 1 f).

WY 5 VYB IY cos (30° 1 f)
Since the load is balanced, the line voltage VRB 5 VYB 5 VL and the line current IR 5 IY 5 IL
WR 5 VL·IL cos (30° – f)
WY 5 VLIL cos (30° 1 f)
Adding WR and WY gives total power in the circuit, thus
WR + WY = 3 VL I L cos f
From the two wattmeter readings, it is clear that for the same load angle f, wattmeter WR registers more
power when the load is inductive. It is also connected in the leading phase as the phase sequence is RYB.
Therefore, WR is higher reading wattmeter in the circuit of Fig. 9.46. In other words, if the load is capacitive,
the wattmeter connected in the leading phase reads less for the same load angle. So, if we know the nature
of the load, we can easily identify the phase sequence of the system. The higher reading wattmeter always
reads positive. By proper manipulation of two wattmeter readings, we can obtain the power factor of the load.
WR 5 VLIL cos (30° – f)
(Higher reading)
WY 5 VLIL cos (30° 1 f)
(Lower reading)
WR + WY = 3 VL I L cos f
WR – WY 5 VLIL sin f
Taking the ratio of the above two values, we get
WR −WY
tan f
=
WR + WY
3
or
 W −WY 

tan f = 3  R
W + W 
Y 
 R
 W −WY
f = tan−1 3  R
W + W
Y
 R
Thereafter, we can find cos f.




EXAMPLE 9.22
The two-wattmeter method is used to measure power in a three-phase load. The wattmeter readings are 400
W and –35 W. Calculate (a) total active power, (b) power factor, and (c) reactive power.
Solution From the given data, the two wattmeter readings WR 5 400 W (Higher reading wattmeter) WY 5 –35
W (Lower reading wattmeter).
(a) Total active power 5 W1 1 W2
5 400 1 (–35) 5 365 W
(b) tan f = 3
400 − (−35)
WR WY
435
= 3
= 3×
= 2.064
400 + (−35)
WR WY
365
376
Circuits and Networks
f 5 tan–1 2.064 5 64.15°; P.F 5 0.43
5
(c) Reactive power
3 VL I L sin f
We know that WR – WY 5 VLIL sin f

WR – WY 5 400 – (– 35) 5 435
Reactive power 5
3 3 435 5 753.44 VAR
EXAMPLE 9.23
The input power to a three-phase load is 10 kW at 0.8 Pf. Two wattmeters are connected to measure the
power, find the individual readings of the wattmeters.
Solution Let WR be the higher reading wattmeter and WY the lower reading wattmeter
WR 1 WY 5 10 kW
f5
cos–1
tan f = 0.75 = 3
or
WR −WY =
(0.75)
3
(9.1)
0.8 5 36.8°
WR −WY
WR + WY
(WR + WY ) =
0.75
3
×10 kW
5 4.33 kW
(9.2)
From Eqs (9.1) and (9.2), we get
WR 1 WY 5 10 kW
WR −WY − 4.33 kW
2WR −14.33 kW
or
WR 5 7.165 kW
WY 5 2.835 kW
9.11.5 Variation in Wattmeter Readings with Load Power Factor
It is useful to study the effect of the power factor on the readings of the wattmeter. In Section 9.11.4, we have
proved that the readings of the two wattmeters depend on the load power factor angle f, such that
WR 5 VLIL cos (30 – f)°
WY 5 VLIL cos (30 1 f)°
We can, therefore, make the following deductions
1. When f is zero, i.e. power factor is unity, from the above expressions we can conclude that the two
wattmeters indicate equal and positive values.
2. When f rises from 0 to 60°, i.e. upto power factor 0.5, the wattmeter WR reads positive (since it
is connected in the leading phase); whereas wattmeter WY reads positive, but less than WR. When
f 5 60°, WY 5 0 and the total power is being measured only by wattmeter WR.
3. If the power factor is further reduced from 0.5, i.e. when f is greater than 60°, WR indicates positive
value, whereas WY reads down scale reading in such case. As already explained in Section 9.11.3,
377
Polyphase Circuits
the connections of either the current coil, or the pressure coil of the corresponding wattmeter have to
be interchanged to obtain an upscale reading, and the reading thus obtained must be given a negative
sign. Then the total power in the circuit would be WR 1 (– WY) 5 WR – WY. Wattmeter WY reads
downscale for the phase angle between 60° and 90°. When the power factor is zero (i.e. f 5 90°),
the two wattmeters will read equal and opposite values.
i.e.
WR 5 VLIL cos (30 – 90)° 5 0.5 VLIL
WY 5 VLIL cos (30 1 90)° 5 –0.5 VLIL
9.11.6 Leading Power Factor Load
The above observations are made considering the lagging power
factor. Suppose the load in Fig. 9.46 (a) is capacitive, the wattmeter
connected in the leading phase would read less value. In that case,
WR will be the lower reading wattmeter, and WY will be the higher
reading wattmeter. Figure 9.48 shows the phasor diagram for the
leading power factor.
As the power factor is leading, the phase currents, IR, IY and IB
are leading their respective phase voltage by an angle f. From Fig.
9.48, the reading of the wattmeter connected in the leading phase
is given by
WR 5 VRB . IR cos (30 1 f)°
5 VLIL cos (30 1 f)° (lower reading wattmeter)
Similarly, the reading of the wattmeter connected in the lagging
phase is given by
Fig. 9.48
WY 5 VYB IY cos (30 – f)°
5 VLIL cos (30 – f)° (higher reading wattmeter)
Again, the total power is given by
WR + WY = 3 VL I L cos f
WY – WR 5 VLIL sin f
WY −WR
WY + WR
A comparison of this expression with that of lagging power factor reveals the fact that the two wattmeter
readings are interchanged, i.e. for lagging power factor, WR is the higher reading wattmeter, and WY is the lower
reading wattmeter; where as for leading power factor, WR is the lower reading wattmeter and WY is the higher
reading wattmeter. While using the expression for power factor, whatever may be the nature of the load, the lower
reading is to be subtracted from the higher reading in the numerator. The variation in the wattmeter reading with
the capacitive load follows the same sequence as in inductive load, with a change in the roles of wattmeters.
Hence tan f = 3
EXAMPLE 9.24
The readings of the two wattmeters used to measure power in a capacitive load are –3000 W and 8000 W,
respectively. Calculate (a) the input power, and (b) the power factor at the load. Assume RYB sequence.
Solution (a) Total power 5 WR 1 WY 5 –3000 1 8000 5 5000 W
378
Circuits and Networks
(b) As the load is capacitive, the wattmeter connected in the leading phase gives less value.
WR 5 –3000

Consequently, WY 5 8000
tan f = 3
(8000 −(−3000))
WY −WR
= 3
= 3.81
5000
WY + WR
f 5 75.29° (lead); cos f 5 0.25

9.11.7 Reactive Power with Wattmeter
We have already seen in the preceding section that the difference between higher reading wattmeter and
lower reading wattmeter yields VL IL sin f. So, the total reactive power 5 3 VL I L sin f . Reactive power in
a balanced three-phase load can also be calculated by using a single wattmeter.
As shown in Fig. 9.49 (a), the current coil of the wattmeters is connected in any one line (R in this case),
and the pressure coil across the other two lines (between Y and B in this case). Assuming phase sequence RYB
and an inductive load of angle f, the phasor diagram for the circuit in Fig. 9.49 (a) is shown in Fig. 9.49 (b).
Fig. 9.49
From Fig. 9.49 (a), it is clear that the wattmeter power is proportional to the product of current through its
current coil, IR, voltage across its pressure coil, VYB, and cosine of the angle between VYB and IR.
VYB 5 VYN – VBN 5 VL
or
From the vector diagram the angle between VYB and IR is (90 – f)°

wattmeter reading 5 VYB IR cos (90 – f)°
5 VL IL sin f VAR
If the above expression is multiplied by
3 , we get the total reactive power in the load.
Polyphase Circuits
379
EXAMPLE 9.25
A single wattmeter is connected to measure reactive power of a three-phase, three-wire balanced load as
shown in Fig. 9.49 (a). The line current is 17 A and the line voltage is 440 V. Calculate the power factor of
the load if the reading of the wattmeter is 4488 VAR.
Solution We know that wattmeter reading is equal to VL IL sin f.

4488 5 440 3 17 sin f
sin f 5 0.6
Power factor 5 cos f 5 0.8
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 9
In the two-wattmeter method of power measurement, the power registered by one wattmeter is 3500
W, while the other reads down scale. After reversing the later, it reads 300 W. Determine the total
power in the circuit and the power factor.
rrr 9-9.2 Three impedances 20 0° V; 16 20° V and 25 90° V are connected in delta across a balanced
supply of 250 0° V. Find phase currents; active power, and Reactive power in each phase and
total apparent power.
rrr 9-9.1
Frequently Asked Questions linked to LO9
rrr 9-9.1
rrr9-9.2
rrr 9-9.3
rrr9-9.4
9.12
The two-wattmeter method produces wattmeter readings P1 = 1560 W and P2 = 2100 W when
connected to a delta-connected load. If the line voltage is 220 V, calculate (a) the per-phase average
power, (b) the per-phase reactive power, (c) the power factor, and (d) the phase impedance.
[AU May/June 2013]
The two-wattmeter method is used to measure power in a three-phase delta-connected load. The
dlta-connected load consists of ZRY = 10 j 10 , ZYB = 15 – j15 , and ZBR = 20 + j 10 and it is
connected to a 400 V, three-phase supply of phase sequence RYB. Calculate the readings of the
wattmeter with current coil in lines R and B.
[AU May/June 2014]
Show that two wattmeters are sufficient to measure power in a balanced or un-balanced threephase load connected to a balanced supply.
[AU April/May 2011]
A three-phase, 220 V, 50 Hz, 11.2 kW induction motor has a full-load efficiency of 88% and draws
a line current of 38 A under full load, when connected to a three-phase, 220 V supply. Find the
readings on two wattmeters connected in the circuit to measure the input to the motor. Determine
also the power factor at which the motor is operating.
[AU April/May 2011]
EFFECTS OF HARMONICS
The relationship between line and phase quantities for wye and delta
LO 10
connections as derived earlier are strictly valid only if the source voltage is
purely sinusoidal. Such a waveform is an ideal one. Modern alternations are
designed to give a terminal voltage which is almost sinusoidal. But it is nearly
impossible to realise an ideal waveform in practice. All sinusoidally varying alternating waveforms deviate
to a greater or lesser degree, from an ideal sinusoidal shape. Due to non-uniform distribution of the field flux
and armature reaction in ac. machines, the current and voltage waves may get distorted. Such waveforms are
referred to as non-sinusoidal or complex waveforms. In modern machines this distortion is relatively small.
All non-sinusoidal waves can be broken up into a series of sinusoidal waves whose frequencies are integral
380
Circuits and Networks
multiples of the frequency of the fundamental wave. The sinusoidal components of a complex wave are called
harmonics. It is therefore necessary to consider the effect of certain harmonics on currents and voltages in the
phase of three-phase wye and delta systems in effecting the line and phase quantities.
The fundamental wave is called the basic wave or first harmonic. The second harmonic has a frequency
of twice the fundamental, the third harmonic frequency is three times the fundamental frequency, and so on.
Each harmonic is a pure sinusoid. Waves having 2f, 4f, 6f, etc. are called even harmonics and those having
frequencies 3f, 5f, 7f, etc. are called odd harmonics. Since the negative half of the wave is a reproduction of
the positive half, the even harmonics are absent. Therefore, a complex wave can be represented as a sum of
fundamental and odd harmonics.
9.12.1 Harmonic Effect in Wye
Let us consider a wye connected generator winding, whose arrangement is shown diagrammatically in Fig. 9.50.
The voltage induced in phase a of the three-phase symmetrical
system, including odd harmonics is given by
Vna 5 Em1sin(vt 1 u1) 1 Em3 sin(3vt 1 u3)
1 Em5 sin(5vt 1 u5) 1 …
(9.3)
where Em1, Em3, Em5, etc. are the peak values of the fundamental
and other harmonics and u1, u3, u5, etc. are phase angles.
Assuming abc phase sequence. The voltage in phase b will be
Vnb 5 Em1 sin (vt 1 u1 – 120°)
1 Em3 sin (3vt – 360° 1 u3)
1 Em5 sin (5vt – 600° 1 u5) 1 …
Fig. 9.50
5 Em1 sin (vt 1 u1 – 120°) 1 Em3 sin (3vt 1 u3)
1 Em5 sin (5vt 1 u5 – 240°) 1 …
(9.4)
The voltage in phase c will be
Vnc 5 Em1 sin (vt 1 u1 – 240°) 1 Em3 sin (3vt 1 u3 – 720°)
1 Em5 sin (5vt 1 u5 – 1200°) 1 …
5 Em1 sin (vt 1 u1 – 240°) 1 Em3 sin (3vt 1 u3)
1 Em5 sin (5vt 1 u5 – 120°) 1 …
(9.5)
Equations (9.3), (9.4), and (9.5) show that all third harmonics are in time phase with each other in all the
three phases as shown in Fig. 9.51 (a). The same applies to the, ninth, fifteenth, twenty first… harmonics, i.e.
all odd multiples of 3. Other than odd multiples of 3, the fifth, seventh, eleventh… and all other harmonics
are displaced 120° in time phase mutually with either the same phase sequence or opposite phase sequence
compared with that of the fundamentals. Fifth harmonics and seventh harmonic sequences are shown in Figs
9.5 (c) and 9.5 (d) respectively.
Summarising the above facts, the fundamental and all those harmonics obtained by adding a multiple of
6, i.e. 1, 7, 13, 19,…, etc. will have the same sequence. Similarly, the fifths and all harmonics obtained by
adding a multiple of 6, i.e. 5, 11, 17, 23,…, etc. will have sequence opposite to that of the fundamental.
Polyphase Circuits
381
Fig. 9.51
Voltage Relations The voltage between lines ab in the wye connected winding in Fig. 9.50 may be
written as
eab 5 ean 1 enb
Adding of each harmonic separately is shown in Fig. 9.52.
Fig. 9.52
It is seen from the vector diagrams of Fig. 9.52 that there is no third harmonic component in the line
voltage. Hence, the rms value of the line voltage is given by
Eab = 3
Em21 + Em2 5 + Em2 7 + …
2
(9.6)
382
Circuits and Networks
and the rms value of the phase voltage is
Ena =
Em21 + Em2 3 + Em2 5 + Em2 7 + …
(9.7)
2
It is seen from the above equations that in a wye-connected system, the line voltage is not equal to
times the phase voltage if harmonics are present. This is true only when the third harmonics are absent.
3
Current Relations Similar to the complex voltage wave, the instantaneous value of the complex current wave can be written as
i 5 Im1 sin (vt 1 f1) 1 Im3 sin (3vt 1 f3) 1 Im5 sin (5vt 1 f5)
(9.8)
where Im1, Im3, Im5, etc. are the peak values of fundamental and other harmonics; (f1 – f1) is the phase
difference between fundamental component of the harmonic voltage and current and (f3 – f3) is the phase
difference between 3rd harmonics and so on. Applying KCL for the three phase wye connected winding in
Fig. 9.50.
ina 1 inb 1 inc 5 0
(9.9)
The equations for ina, inb and inc can be obtained by replacing currents in the place of voltages in Eqs
(9.3), (9.4) and (9.5) under balanced conditions the sum of the three currents is equal to zero, only when
they have equal magnitudes and displaced by 120° apart in time phase in the three phases. All harmonics
fulfil the above condition except the third harmonics and their odd multiples as they are in the same phase
as shown in Fig. 9.51 (a) or 9.52 (b). Hence, the resultant of ina 1 inb 1 inc consists of the arithmetic sum
of the third harmonics in the three phases. Hence, there must be a neutral wire or fourth wire to provide a
return path for the third harmonic. We can summarise the above facts as follows. In a balanced three-wire
wye connection, all harmonics are present except third and its odd multiples. In a four-wire wye connection,
i.e. with a neutral wire, all harmonics will exist.
9.12.2 Harmonic Effect in Delta
Let the three windings of the generator be delta-connected as shown in Fig. 9.53. Let vna, vnb and vnc be the
phase emfs and vna, vnb and vnc be the terminal voltages of
the three phases a, b and c respectively. According to KVL,
the algebraic sum of the three terminal voltages in the closed
loop is given by
vna 1 vnb 1 vnc
5 vca 1 vab 1 vbc 5 0
(9.10)
Fig. 9.53
There will be a circulating current in the closed loop due to
the resultant third harmonic and their multiple induced emfs.
This resultant emf is dropped in the closed loop impedance.
Hence, the third harmonic voltage does not appear across the
terminals of the delta. Hence, the terminal voltages in delta
connection vca, vab and vbc are given by equations 1, 2, and 3
respectively without third harmonic terms.
383
Polyphase Circuits
Current Relations The three phase windings in Fig. 9.53, carry all the harmonics internally and are given
by
ina 5 ica 5 Im1 sin (vt 1 u1) 1 Im2 sin (3vt 1 u3) 1 Im5 sin (5vt 1 u5) 1 …
(9.11)
inb 5 iab 5 Im1 sin (vt 1 u1 – 120°) 1 Im3 sin (3vt 1 u3 – 360°)
1 Im5 sin (5vt 1 u5 – 600°) 1 …
5 Im1 sin (vt 1 u1 – 120°) 1 Im3 sin (3vt 1 u3) 1 Im5 sin (5vt 1 u5 – 240°) 1 …
(9.12)
inc 5 ibc 5 Im1 sin (vt 1 u1 – 240°) 1 Im3 sin (3vt 1 u3 – 720°) 1 Im5 sin (5vt 1 u5 – 1200°)
5 Im1 sin (vt 1 u1 – 240°) 1 Im3 sin (3vt 1 u3) 1 Im5 sin (5vt 1 u5 – 120°)
(9.13)
Equations (9.11), (9.12) and (9.13) represent the phase currents in the delta connection. The line currents
IAa IBb and ICc can be obtained by applying KCL at the three junctions of the delta in Fig. 9.53. The current
vector diagrams are similar to the voltage vector diagrams shown in Fig. 9.52 except that the voltages are to
be replaced with currents. The line currents can be obtained in terms of phase currents by applying KCL at
three junctions as follows:
iAa 5 iab – iCa
5 Im1 sin (vt 1 u1 – 120°) 1 Im5 sin (5vt 1 u5 – 240°)
– Im1 sin (vt 1 u1) – Im5 sin (5vt 1 u5)
(9.14)
iBb 5 ibc – iab
5 Im1 sin (vt 1 u1 – 240°) 1 Im5 sin (5vt 1 u5 – 120°)
– Im1 sin (vt 1 u1 – 120°) – Im5 sin (5vt 1 u5 – 240°)
(9.15)
iCc 5 ica – ibc
5 Im1 sin (vt 1 u1) 1 Im5 sin (5vt 1 u5) – Im1 sin (vt 1 u1 – 240°)
– Im5 sin (5vt 1 u5 – 120°)
(9.16)
Equations (9.14), (9.15) and (9.16) indicate that no third harmonic currents can exist in the line currents
of a delta connection.
The rms value of the line current from the above equation is
IL = 3
I m21 + I m2 5 + …
(9.17)
2
The rms value of the phase current from Eqs (9.11), (9.12) and (9.13) is given by
I ph =
I m21 + I m2 3 + I m2 5 + …
(9.18)
2
It is seen from Eqs (9.17) and (9.18), that in a delta-connected system the line current is not equal to
times the phase current. It is only true when there are no third harmonic currents in the system.
9.13
EFFECTS OF PHASE-SEQUENCE
The effects of phase sequence of the source voltages are not of considerable
importance for applications like lighting, heating, etc. but in case of a three-phase
induction motor, reversal of sequence results in the reversal of its direction. In
the case of an unbalanced polyphase system, a reversal of the voltage phase
LO 11
3
384
Circuits and Networks
sequence will, in general, cause certain branch currents to change in magnitude as well as in phase position.
Even though the system is balanced, the readings of the two-wattmeters in the two wattmeter method of
measuring power interchange when subjected to a reversal of phase sequence when two or more three-phase
generators are running parallel to supply a common load, reversing the phase sequence of any one machine
cause severe damage to the entire system. Hence, when working on such systems, it is very important to
consider the phase sequence of the system. Unless otherwise stated, the term “phase sequence” refers to
voltage phase sequence. The line currents and phase currents follow the same sequence as the system voltage.
The phase sequence of a given system, is a small meter with three long connecting leads in side which it has a
circular disc. The rotation of which previously been checked against a known phase sequence. In three-phase
systems, only two different phase sequences are possible. The three leads are connected to the three lines
whose sequence is to be determined, the rotation of the disc can be used as an indicator of phase sequence.
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 11
rrr 9-11.1
9.14
The power taken by a 440 V, 50 Hz, three-phase induction motor on full load is measured by two
wattmeters, which indicate 250 W and 1000 W, respectively. Calculate (a) the input, (b) the
power factor, (c) the current, and (d) the motor output, if the efficiency is 80%.
POWER FACTOR OF AN UNBALANCED SYSTEM
The concept of power factor in three-phase balanced circuits has been discussed in
LO 12
Section 9.11.4. It is the ratio of the phase watts to the phase volt-amperes of any
one of the three phases. We cannot strictly define the power factor in three-phase
unbalanced circuits, as each phase has a separate power factor. Generally, for
three-phase unbalanced loads, the ratio of total watts 3VL I L cosu to total volt-amperes
good general indication of the power factor.
(
)
(
)
3VL I L is a
Another recognised definition for an unbalanced polyphase system is called the vector power factor, given
by
Power factor =
∑ VI cosu
∑ VI
where ∑VIcosu is the algebraic sum of the active powers of all individual phases given by
∑VIcosu 5 Va Ia cos ua 1 Vb Ib cos ub 1 Vc Ic cos uc 1 …
and ∑ VI =
2
2
(∑ VI cos u) + (∑ VI sin u)
∑VI sin u is the algebraic sum of the individual phase reactive volt-amperes. The following example
illustrates the application of vector power factor for unbalanced loads.
Consider Example 9.19 where the phase voltage and currents have been already calculated. Here VRN 5 230.94
∠0° V, VYN 5 230.94 ∠–120° V, VBN 5 230.94 ∠–240° V
IR 5 24.83 ∠– 63.4° A; IY 5 46.188 ∠– 173.1° A; IB 5 9.23 ∠– 293.13° A
Active power of phase R 5 230.94 3 25.83 3 cos 63.4° 5 2.6709 kW
Active power of phase Y 5 230.94 3 46.188 3 cos 53.1° 5 6.4044 kW
385
Polyphase Circuits
Active power of phase B 5 230.94 3 9.23 3 cos 53.13°
5 1.2789 kW
10.3542 kW
∑VI cosu 5 10.3542 kW
Reactive power of phase R 5 230.94 3 25.83 3 sin 63.4° 5 5.3197 KVAR
Reactive power of phase Y 5 230.94 3 46.188 3 sin 53.1° 5 8.5299 KVAR
Reactive power of phase B 5 230.94 3 9.23 3 sin 53.13° 5 1.7052 KVAR
15.5548 KVAR
∑VI sin u 5 15.5548 KVAR
Power factor =
10.3542
(15.5548) + (10.3542)
2
2
= 0.5541
Additional Solved Problems
PROBLEM 9.1
The phase voltage of a star-connected three-phase ac generator is 230 V. Calculate the (a) line voltage,
(b) active power output if the line current of the system is 15 A at a power factor of 0.7, and (c) active and
reactive components of the phase currents.
Solution The supply voltage (generator) is always assumed to be balanced
 VPh 5 230 V; IL 5 IPh 5 15 A, cos f 5 0.7, sin f 5 0.71
(a) In a star-connected system VL 5
(b) Power output 5
5
3 VPh 5 398.37 V
3 VL IL cos f
3 3 398.37 3 15 3 0.7 5 7244.96 W
(c) Active component of the current 5 IPh cos f
5 15 3 0.7 5 10.5 A
Reactive component of the current 5 IPh sin f
5 15 3 0.71 5 10.65 A
PROBLEM 9.2
A three-phase delta-connected RYB system with an effective voltage of 400 V, has a balanced load with
impedances 3 1 j4 V. Calculate the (a) phase currents, (b) line currents, and (c) power in each phase.
Solution
VL 5 VPh 5 400 V
Assuming RYB phase sequence, we have
VRY 5 400 ∠0°; VYB 5 400 ∠–120°; VBR 5 400 ∠– 240°
Z 5 3 1 j4 5 5 ∠53.1°
386
Circuits and Networks
Fig. 9.54
(a) The three phase currents are
IR =
VRY
400 ∠0°
=
= 80 ∠− 53.1°
Z
5 ∠53.1°
IY =
VYB 400 ∠−120°
=
= 80 ∠ −173.1°
Z
5 ∠53.1°
IB =
VYB 400 ∠− 240°
=
= 80 ∠− 293.1°
5 ∠53.1°
Z
IR 5 80 ∠– 53.1° 5 48.03 – j63.97
IY 5 80 ∠– 173.1° 5 – 79.42 – j9.61
IB 5 80 ∠– 293.1° 5 31.38 1 j73.58
(b) The three line currents are
I1 5 IR – IB 5 138.55 ∠– 83.09°
I2 5 IY – IR 5 138.55 ∠156.9°
I3 5 IB – IY 5 138.55 ∠36.89°
cos f =
RPh 3
= = 0.6
Z Ph 5
(c) Power consumed in each phase 5 VPh IPh cos f
5 400 3 80 3 0.6 5 19200 W
Total power 5 3 3 19200 5 57600 W
PROBLEM 9.3
The load in Problem 9.2 is connected in star with the same phase sequence across the same system.
Calculate (a) the phase and line currents (b) the total power in the circuit, and (c) phasor sum of the three
line currents.
Solution The circuit is shown in Fig. 9.55.
Polyphase Circuits
387
Fig. 9.55
Assuming RYB phase sequence, since
VL = 400 V
VPh =
400
3
= 230.94 V
Taking VRN as reference, the three phase voltages are VRN 5 230.94∠0°; VYN 5 230.94 ∠–120°; and
VBN 5 230.9 ∠–240°.
The three line voltages, VRY, VYB and VBR are 30° ahead of their respective phase voltages.
IPh 5 IL; ZPh 5 3 1 j4 5 5 ∠53.1°
(a) The three phase currents are
IR =
VRN
230.94 ∠0°
=
= 46.18 ∠− 53.1°
Z Ph
5 ∠53.1°
IY =
VYN
230.09 ∠−120°
=
= 46.18 ∠−173.1°
5 ∠53.1°
Z Ph
IB =
VBN
230.09 ∠− 240°
=
= 46.18 ∠− 293.1°
Z Ph
5 ∠53.1°
(b) Total power 5
5
3 VL IL cos f
3 3 400 3 46.18 3 0.6 5 19196.6 W
Thus, it can be observed that the power consumed in a delta load will be three times more than that in the
star connection
(c) Phasor sum of the three line currents
5 IR 1 IY 1 IB
5 46.18 ∠–53.1° 1 46.18 ∠–173.1° 1 46.18 ∠–293.1° 5 0
388
Circuits and Networks
PROBLEM 9.4
A three-phase balanced delta-connected load with line voltage of 200 V, has line currents as I1 5 10 ∠90°;
I2 5 10 ∠–150° and I3 5 10 ∠–30°. (a) What is the phase sequence? (b) What are the impedances?
Solution Figure 9.56 (a) represents all the three line currents in the phasor diagram.
(a) It can be observed from Figs 9.56 (a) and (b) that the current flowing in the
line B, i.e. I3 lags behind I1 by 120°, and the current flowing in line Y, i.e. I2
lags behind i3 by 120°.  The phase sequence is RBY.
(b) VPh 5 VL 5 200
Fig. 9.56 (a)
I Ph =
IL
Z Ph =
VPh 200 3
=
= 34.64 V
I Ph
10
3
=
10
3
Fig. 9.56 (b)
PROBLEM 9.5
Three equal inductors connected in star take 5 kW at 0.7 Pf when connected to a 400 V, 50 Hz three-phase,
three-wire supply. Calculate the line currents (a) if one of the inductors is disconnected, and (b) If one of the
inductors is short circuited.
Solution Total power when they are connected to the 400 V supply
P5
3 VL IL cos f 5 5000 W
I Ph = I L =
5000
3 × 400 × 0.7
Impedance/phase =
Fig. 9.57 (a)
= 10.31 A
VPh
400
=
= 22.4 V
I Ph
3 ×10.31
RPh 5 ZPh cos f 5 22.4 3 0.7 5 15.68 V
XPh 5 ZPh sin f 5 22.4 3 0.714 5 16 V
389
Polyphase Circuits
(a) If the phase Y is disconnected from the circuit, the other two inductors are connected in series
across the line voltage of 400 V as shown in Fig. 9.57 (a).
400
IR = IB =
= 8.928 A
2× Z Ph
IY = 0
(b) If phases Y and N are short circuited as shown in Fig. 9.57 (b), the phase voltages VRN and VBN will
be equal to the line voltage 400 V.
I Ph = I R = I B =
400 400
=
= 17.85 A
Z Ph 22.4
The current in the Y phase is equal to the phasor sum of the
R and B.
\
Fig. 9.57 (b)
 60 
IY = 2× I Ph cos   = 30.91 A
 2 
PROBLEM 9.6
For the circuit shown in Fig. 9.58, calculate the line current,
the power and the power factor. The value of R, L, and C
in each phase are 10 ohms, 1 H and 100 mF, respectively.
Solution
Let us assume RYB sequence.
VRN =
400
3
∠0° = 231∠0°;VYN = 231∠−120°;VBN = 231∠− 240°
Admittance of each phase YPh =
1
1
+
+ j vC
R j vL
1
1
+
+ j 314 ×100 ×10−6
10 j 314
YPh 5 0.1 1 j28.22 3 10–3
= 0.103 ∠15.75° �
IPh 5 VPh YPh
5 (231 ∠0°) (0.103 ∠15.75°)
5 23.8 ∠15.75° A
Power 5 3 VL IL cos f
=
5 3 3 400 3 23.8 cos 15.75°
5 15869.57 W
Power factor 5 cos 15.75° 5 0.96 leading
Fig. 9.58
390
Circuits and Networks
PROBLEM 9.7
For the circuit shown in Fig. 9.59, an impedance is
connected across YB, and a coil of resistance 3 V and
inductive reactance of 4 V is connected across RY.
Find the value of R and X of the impedance across
YB such that I2 5 0. Assume a balanced three-phase
supply with RYB sequence.
Solution As usual IR, IY, and IB are phase currents,
and I1, I2, and I3 are line currents.
Applying KCL at the node Y, we have
I 2 = IY − I R
Fig. 9.59
I 2 = IY − I R
Since
I2 = 0
IY = I R

IR =
VRY
V
, IY = YB
ZYB
3 + j4
VRY = V ∠0°,VYB = V ∠−120°
V ∠0° V ∠−120°
=
ZYB
3 + j4
V ∠−120°
ZYB =
(3 + j 4)
V ∠0°
= 1.96 − j 4.6
R 5 1.96 V; X 5 4.6 V (capacitive reactance)

PROBLEM 9.8
A symmetrical three-phase 440 V system supplies a balanced delta-connected load. The branch current is
10 A at a phase angle of 30°, lagging. Find (a) the line current, (b) the total active power, and (c) the total
reactive power. Draw the phasor diagram.
Solution (a) In a balanced delta-connected system
IL 5 3 IPh 5 3 3 10 5 17.32 A
(b) Total active power
5
3 VL IL cos f
5 3 3 440 3 17.32 3 cos 30° 5 11.431 kW
(c) Total reactive power
5 3 VL IL sin f
Polyphase Circuits
5
391
3 3 440 3 17.32 3 sin 30° 5 6.5998 KVAR
The phasor diagram is as under.
Fig. 9.60
VR, VY, and VB are phase voltages, and are equal to the line values. IR, IY, and IB are the phase currents, and
lag behind their respective phase voltages by 30°. Line currents (IR – IB), (IY – IR) and (IB – IY) lag behind their
respective phase currents by 30°.
PROBLEM 9.9
Find the line currents and the total power consumed by the unbalanced delta-connected load shown in Fig.
9.61.
Fig. 9.61
Solution
Assuming RYB phase sequence, from the given data,
IR 5 10 ∠–36.88°; IY 5 5 ∠45.57°; IB 5 7 ∠0°
Line currents are
I1 5 IR – IB 5 6.08 ∠–80°
I2 5 IY – IR 5 10.57 ∠11.518°
I3 5 IB – IY 5 5 ∠– 45.56°
Total power is the sum of the powers consumed in all the three phases.

Power in RY 5 VRY 3 IR 3 0.8
5 400 3 10 3 0.8 5 3200 W
392
Circuits and Networks
Power in YB 5 VYB 3 IY 3 0.7
5 400 3 5 3 0.7 5 1400 W
Power in BY 5 VBR 3 IB 3 1
5 400 3 7 3 1 5 2800 W
Total power 5 3200 1 1400 1 2800 5 7400 W
PROBLEM 9.10
A delta-connected three-phase load has 10 V between R and Y, 6.36 mH between Y and B, and 636 mF
between B and R. The supply voltage is 400 V, 50 Hz. Calculate the line currents for RBY phase sequence.
Solution ZRY 5 10 1 j0 5 10 ∠0°; ZYB 5 0 1 jXL 5 0 1 jXL
Z BR
5 0 1 j2p f L 5 2 ∠90°3
j
= 0 − jX C = 0 −
= 5 ∠− 90°
2pfC
Fig. 9.62
Since the phase sequence is RBY, taking VRY as reference voltage, we have
VRY 5 400 ∠0°; VBR 5 400 ∠–120°; VYB 5 400 ∠–240°
IR =
VRY
400 ∠− 0°
=
= 40 ∠0°
10 ∠0°
10 ∠0°
IY =
VYB
400 ∠− 240°
=
= 200 ∠− 330°
2 ∠90°
2 ∠90°
VBR
400 ∠−120°
=
= 80 ∠− 30°
5 ∠− 90°
5 ∠− 90°
The three line currents are
I1 5 IR – IB 5 40 ∠0° – 80 ∠–30° 5 49.57 ∠126.2°
IB =
I2 5 IY – IR 5 200 ∠–300° ∠– 40 ∠0° 5 166.56 ∠36.89°
I3 5 IB – IY 5 80 ∠– 30° – 200 ∠–330° 5 174.35 ∠233.41°
Polyphase Circuits
393
PROBLEM 9.11
The power consumed in a three-phase balanced star-connected load is 2 kW at a power factor of 0.8 lagging.
The supply voltage is 400 V, 50 Hz. Calculate the resistance and reactance of each phase.
400
Solution Phase voltage =
3
3 VL IL cos f
Power consumed 5 2000 W 5
2000
Phase current or line current I L =
3 × 400 × 0.8
= 3.6 A
Impedance of each phase
Z Ph =
VPh
400
=
= 64.15 A
I Ph
3 ×3.6
Since the power factor of the load is lagging, the reactance is
inductive reactance. From the impedance triangle shown in Fig.
9.63, we have
Resistance of each phase RPh 5 ZPh cos f
5 64.15 3 0.8 5 51.32 V
Reactance of each phase XPh 5 ZPh sin f
Fig. 9.63
5 64.15 3 0.6 5 38.5 V
PROBLEM 9.12
A symmetrical three-phase 100 V; three-wire supply feeds an unbalanced star-connected load, with
impedances of the load as ZR 5 5 ∠0° V, ZY 5 2 ∠90° V and ZB 5 4 ∠–90° V. Find the (a) line currents,
(b) voltage across the impedances, and (c) the displacement neutral voltage.
Solution As explained earlier, this type of unbalanced Y-connected three-wire load can be solved either by
star-delta conversion method or by applying Millman’s theorem.
(a) Star-Delta Conversion Method
The unbalanced star-connected load and its equivalent delta load are shown in Figs 9.64 (a) and (b).
ZR ZY 1 ZY ZB 1 ZB ZR 5 (5 ∠0°) (2 ∠90°) 1 (2 ∠90°) (4 ∠–90°)
1 (4 ∠–90°) (5 ∠0°)5 8 – j10
5 12.8 ∠–51.34°
Z RY =
Z R ZY + ZY Z B + Z B Z R 12.8 ∠−51.34°
=
= 3.2 ∠38.66°
ZB
4 ∠−90°
394
Circuits and Networks
Fig. 9.64
ZYB =
Z R ZY + ZY Z B + Z B Z R 12.8 ∠−51.34
=
ZR
5 ∠0
Z BR =
Z R ZY + ZY Z B + Z B Z R 12.8 ∠−51.34°
=
= 6.4 ∠−141.34°
2 ∠90°
ZY
= 2.56 ∠− 51.34
Taking VRY as the reference, we have
VRY 5 100 ∠0°; VYB 5 100 ∠–120°; VBR 5 100 ∠–240°
The three phase currents in the equivalent delta load are
IR =
VRY
100 ∠0°
=
= 31.25 ∠−38.66°
Z RY
3.2 ∠38.66°
IY =
VYB
100 ∠−120°
= 39.06 ∠−68.66°
=
ZYB 2.56 ∠−51.34°
VBR
100 ∠−240°
=
= 15.62 ∠−98.66°
Z BR 6.4 ∠−141.34°
The line currents are
IB =
I1 5 IR – IB 5 31.25 ∠–38.66° –15.62 ∠–98.66°
5 (24.4 – j19.52) – (– 2.35 1 j15.44) 5 (26.75 – j4.08)
5 27.06 ∠–8.671°
I2 5 IY – IR 5 39.06 ∠–68.66° – 31.25 ∠–38.66°
5 (14.21 – j36.38) – (24.4 – j19.52) 5 (–10.19 – j16.86)
5 19.7 ∠238.85°
I3 5 IB – IY 5 15.62 ∠–98.66° – 39.06 ∠–98.66°
5 (–2.35 – j15.44) – (14.21 – j36.38) 5 (–16.56 1 j20.94)
5 26.7 ∠128.33°
Polyphase Circuits
These line currents are also equal to the line (phase) currents of the original star connected load.
(b) Voltage drop across each star-connected load will be as under.
Voltage across ZR 5 I1 3 ZR
5 (27.06 ∠–8.671°) (5 ∠0°) 5 135.3° ∠–8.67°
Voltage across ZY 5 I2 3 ZY
5 (19.7 ∠238.85°) (2 ∠90°) 5 39.4° ∠328.85°
Voltage across ZB 5 I3 3 ZB
5 (26.7 ∠128.33°) (4 ∠–90°) 5 106.8° ∠38.33°
(c) By Applying Millman’s Theorem
Consider Fig. 9.64 (c), taking VRY as reference line voltage 5 100 ∠0°.
Fig. 9.64 (c)
Phase voltages log 30° behind their respective line voltages. Therefore, the three phase voltages are
VRO =
VYO =
VBO =
100
3
100
3
100
∠− 30°
∠−150°
∠− 270°
3
1
1
YR =
=
= 0.2 ∠0°
Z R 5 ∠0°
1
1
YY =
=
= 0.5 ∠−90°
ZY
2 ∠90°
1
1
YB =
=
= 0.2
25 ∠90°
Z B 4 ∠− 90°
VROYR 1 VYOYY 1 YBOYB 5 (57.73 ∠–30°) (0.2 ∠0°)
395
396
Circuits and Networks
1 (57.73 ∠–150°) (0.5 ∠– 90°)
1 (57.73 ∠–270°) (0.25 ∠90°)
5 11.54 ∠–30° 1 28.86 ∠–240° 1 14.43 ∠–180°
5 (10 – j5.77) 1 (–14.43 1 j25) 1 (–14.43 1 j0)
5 –18.86 1 j19.23 5 26.93 ∠134.44°
YR 1 YY 1 YB 5 0.2 1 0.5 ∠–90° 1 0.25 ∠90°
5 0.32 ∠–51.34°
VO ′O =
VRO YR + VYO YY + VBO YB
26.93 ∠134.44°
=
0.32 ∠− 51.34°
YR + YY + YB
5 84.15 ∠185.78°
The three load phase voltages are
VRO9 5 VRO – VO9O
5 57.73 ∠–30° – 84.15 ∠185.78°
5 (50 – j28.86) – (– 83.72 – j8.47)
5 (133.72 – j20.4) 5 135.26 ∠–8.67°
VYO9 5 VYO – VO9O
5 57.73 ∠–150° – 84.15 ∠185.78°
5 (–50 – j28.86) – (– 83.72 – j8.47)
5 33.72 – j20.4 5 39.4 ∠–31.17° or 39.4 ∠328.8°
VBO9 5 VBO – VO9O
5 57.73 ∠–270° –84.15 ∠185.78°
5 0 1 j57.73 1 83.72 1 j8.47
5 83.72 1 j66.2 5 106.73 ∠38.33°
135.26 ∠− 8.67°
= 20.06 ∠− 8.67°
5 ∠0°
39.4 ∠328.80°
IY =
= 19.7 ∠238.8°
2 ∠90°
106.73 ∠38.33°
IB =
4 ∠− 90°
IR =
5 26.68 ∠128.33°
397
Polyphase Circuits
PROBLEM 9.13
A three-phase three-wire unbalanced load is star-connected. The phase voltages of two of the arms are
VR 5 100 ∠–10°; VY 5 150 ∠100°
Calculate voltage between star point of the load and the supply neutral.
Solution As shown in Fig. 9.65,
Fig. 9.65
or
VRO 5 VRO9 1 VO9O
VO9O 5 VRO – VRO9
Also VO9O 5 VYO – VYO9
Let VRO 5 V ∠0°
(9.19)
(9.20)
Assuming RYB phase sequence,
VYO 5 V ∠–120°
Substituting in Eqs (9.19) and (9.20), we have
VO9O 5 V ∠0° – 100 ∠–10°
VO9O 5 V ∠– 120° – 50 ∠100°
Subtracting Eq. (9.22) from Eq. (9.21), we get
O 5 [(V 1 jO) – (98.48 – j17.36)] – [(0.5V 1 j0.866V ) – (–26.04 1 j147.72)]
O 5 1.5V – j0.8666V – 124.52 1 j165.08
5 V (1.5 – j0.866) 5 124.52 – j165.08
V=
124.52 − j165.08° 206.77 ∠− 52.97°
=
1.5 − j 0.866
1.732 ∠− 30°
V 5 119.38 ∠–22.97°
Voltage between O9O 5 VRO – VRO9
VO9O 5 119.38 ∠–22.97° – 100 ∠–10°
5 109.91 – j46.58 – 98.48 1 j17.36
5 11.43 – j29.22 5 31.37 ∠– 68.63°
(9.21)
(9.22)
398
Circuits and Networks
PROBLEM 9.14
Find the reading of a wattmeter in the circuit shown in Fig. 9.66 (a). Assume a symmetrical 400 V supply with
RYB phase sequence and draw the vector diagram.
Fig. 9.66 (a)
Solution The reading in the wattmeter is equal to the product of the current through the current coil I1
voltage across its pressure coil VYB and cos of the angle between the VYB and I1.
IR =
VRY
400 ∠0°
=
= 8 ∠90°
− j 50 50 ∠−90°
IB =
VBR
400 ∠−240°
=
30 + j 40
50 ∠53.113°
= 8 ∠− 293.13° or 8 ∠66.87°
Line current
I1 5 IR – IB
5 8∠90° – 8 ∠–293.13°
5 0 1 j8 – 3.14 – j7.35 5 –3.14
1 j0.65 5 3.2 ∠168.3°
Fig. 9.66 (b)

From the vector diagram in Fig. 9.66 (b), it is clear that
the angle between VYB and I1 is 71.7°.
Wattmeter reading is equal to VYB 3 I1 cos 71.7°
5 400 3 3.2 3 cos 71.7 5 402 W
PROBLEM 9.15
Calculate the total power input and readings of the two wattmeters connected to measure power in a threephase balanced load, if the reactive power input is 15 kVAR, and the load pf is 0.8.
Polyphase Circuits
399
Solution Let W1 be the lower reading wattmeter and W2 the higher reading wattmeter.
cos f = 0.8
f = 36.86°
tan f =
Reactive power
Active power
= 3
or
W2 −W1
W2 + W1
Reactive power = 3 (W2 −W1 ) = 15000
= W2 −W1 = 8660.508 W
\
0.75 = 3
(9.23)
15000
W2 + W1
or total power input W2 1 W1 5 34641.01 W
(9.24)
From Eqs (9.23) and (9.24), we get
W2 5 21650.76 W
W1 5 12990.24 W
PROBLEM 9.16
Two wattmeters are connected to measure power in a three-phase circuit. The reading of one of the meters is
5 kW when the load power factor is unity. If the power factor of the load is changed to 0.707 lagging, without
changing the total input power, calculate the readings of the two wattmeters.
Solution Both wattmeters indicate equal values when the power factor is unity
 W1 1 W2 5 10 kW (Total power input)
(9.25)
Let W2 be the higher reading wattmeter.
Then W1 is the lower reading wattmeter.
cos f = 0.707 \ f = 45°
tan f = 3
\
W2 −W1 =
10
W2 −W1
W −W1
1= 3 2
W2 + W1
10
= 5.773 kW
3
From Eqs (9.25) and (9.26),
W2 5 7.886 kW
W1 5 2.113 kW
(9.26)
PROBLEM 9.17
The line currents in a balanced six-phase mesh connected generator are 35.35 A. What is the magnitude of
the phase current?
400
Circuits and Networks
Solution From Section 9.8.2,
I L = 2 I Ph sin
I Ph =
180°
n
35.35
= 35.35
180°
2 sin
6
PROBLEM 9.18
Find the voltage between the adjacent lines of a balanced six-phase star-connected system with a phase
voltage of 132.8 volts.
Solution
From Section 9.7.2, E L = 2 E Ph sin
E L = 2 ×132.8 × sin
180°
n
180°
= 132.8 V
6
PROBLEM 9.19
In the wye- connected system shown in Fig. 9.67, it is assumed
that only fundamental and third harmonic voltages are present
when the voltages are measured with a voltmeter between na and
ba. They are given by 230 and 340 volts respectively. Calculate
the magnitude of the third harmonic voltages in the system.
Solution Only phase voltage Vna of the system shown in Fig.
9.67 contains third harmonic whereas line voltage Vba contains
only first harmonic. Hence,
Fundamental component of the phase =
Fig. 9.67
340
3
 340 2
 = 99.33 V
Third harmonic component = 220 2 − 
 3 
PROBLEM 9.20
Illustrate the effect of reversal of voltage sequence up on the magnitudes of the currents in the system shown
in Example 9.20.
Solution The line currents for RYB sequence have already been calculated. I1 5 28.41 ∠–69.07°, I2 5 29.85
∠136.58°, and I3 5 13 ∠27.60° A.
Polyphase Circuits
401
If the phase sequence is reversed by RBY then
IR =
VRY
400 ∠0°
=
= 25.52 ∠− 60.13° A
Z RY 15.67 ∠60.13°
IY =
VYB
400 ∠− 240°
=
= 9.12 ∠− 289.83° A
ZYB 43.83 ∠49.83°
IB =
VBR
400 ∠−120°
=
= 5.1∠−180.13° A
Z BR 78.36 ∠60.13°
Various line currents are given by
I1 5 IR – IB 5 25.52 ∠–60.13° – 5.1 ∠–180.13° 5 28.41 ∠–51.189° A
I2 5 IY – IR 5 9.12 ∠–289.83° – 25.52 ∠–60.13° 5 32.175 ∠107.37° A
I3 5 IB – IY 5 5.1 ∠–180.13° – 9.12 ∠–289.83° 5 11.85 ∠46.26° A
From the above calculations, it can be verified that the magnitudes of the line currents are not same when
the phase sequence is changed.
PROBLEM 9.21
A balanced delta load is supplied from a symmetrical 3-phase, 400 V, 50 Hz supply system. The current in
each phase is 20 A and lags behind its phase voltage by an angle of 40°. Calculate
(a) The line current
(b) Total power
(c) Also draw the phasor diagram
(d) The wattmeter readings if two wattmeters are used
Solution
(a) I L = 3 I ph = 3 × 20 = 34.64 A
(b) Total power = 3 VL I L cos f
= 3 × 400 ×34.64 × cos 40°
= 18.384 kW
(c) Phasor diagram is shown in Fig. 9.68 (a).
(d) Total active power = W1+W2
= 3 VL I L cos f
= 3 × 400 ×34.64 × cos 40°
On solving,
= 18.384 kW
W1 −W2 = VL I L sin f
5 400 3 34.64 3 sin 40°
5 8.906 kW
W1 5 13.645 kW
W2 5 4.739 kW
(9.27)
(9.28)
402
Circuits and Networks
Fig. 9.68
PROBLEM 9.22
Three identical impedances are star connected across a balanced 440 V; 50 Hz supply. The three line currents
are I R = 20 −40°, IY = 20 −160° and I B = 20 80° ; Find the values of the elements. Total power and the
readings of wattmeters to measure the power.
Solution
VR ph =
VB ph =
Given
440 0°
3
= 254 0° ; VY
440 −240°
3
ph
=
440 −120°
3
= 254 −240°
I R = 20 −40°; IY = 20 −160°; I B = 20 80°
Fig. 9.69
= 254 −120°
Polyphase Circuits
ZR =
254 0°
= 12.7 40° V = (9.728 + j8.16) V
20 −40°
ZY =
254 −120°
= 12.7 40° V = (9.728 + j8.16) V
20 −160°
ZB =
254 −240°
= 12.7 40° V = (9.728 + j8.16) V
20 80°
403
Power consumed 5 3 3 I 2 Rph
5 3 3 (20)2 3 9.728
5 11673 W
Wattmeter readings
W1 5 VL IL cos (30° – )
W2 5 VL IL cos (30° 1 )
where
is the power factor angle between Vph and Iph i.e,
5 40°
W1 5 440 3 20 3 cos (30-40)° 5 8666.3 W
W2 5 440 3 20 3 cos (30140)° 5 3009.7 W
Total power 5 W1 1 W2
5 8666.3 1 3009.7
PT 5 11676 W
PROBLEM 9.23
A
3-phase,
3-wire,
208 V,
CBA
system
has
a
star-connected
load
with
Z A = 5 0° Ω ; Z B = 5 30° Ω; ZC = 10 −60° Ω. Find line currents and voltage across each load impedance.
Draw the phasor diagram.
Solution
Fig. 9.70
404
Circuits and Networks
By converting Y-network into D,
Z BC =
Z A Z B + Z B ZC + ZC Z A 105.85 −31.81°
=
= 21.17 −31.81° V
ZA
5 0°
ZCA =
Z A Z B + Z B ZC + ZC Z A 105.85 −31.81°
=
= 21.17 −61.81° V
ZB
5 30°
Z BA =
Z A Z B + Z B ZC + ZC Z A 105.85 −31.81°
=
= 10.5 −61.81° V
ZC
10 −60°
Phase sequence is CBA.
\
VCB = 208 0° ; VBA = 208 −120° ; VAC = 208 −240°
Phase currents
IC =
VCB
208 0°
=
= 9.82 31.81°
21.17 −31.81°
Z BC
IB =
VBA
208 −120°
=
= 19.65 −148°
Z AB 10.58 28.19°
IA =
VAC
208 −240°
= 9.825 −179°
=
Z AC
21.17 −61.81°
IA
IC
IB
Fig. 9.71
Line currents are
I1 = I C − I A = 9.82 31.81°− 9.825 −179.19°
= 18.9316.30° A
I 2 = I B − I C = 19.65 −148.19°− 9.82 31.81°
= 29.47 −148.19° A
I 3 = I A − I B = 9.825 −179.19° −19.65 −148.19°
= 12.31 56.06° A
Fig. 9.72
Polyphase Circuits
405
Voltages across each load impedance are
VZC = I1 × ZC = (18.9316.3°)(10 −60°)
= 189.3 −43.7° V
VZB = I 2 × Z B = ( 29.47 −148.19°)(5 30°)
= 147.35 −118.19° V
VZA = I 3 × ZC = (12.31 56.06°)(5 0°)
= 61.55 56.06° V
PROBLEM 9.24
A 3-phase, 3-wire star-connected unbalanced load is connected as shown. Find the line currents by node
voltage method. Assume the line voltage as 415 V and abc phase sequence.
Solution
Fig. 9.73
Taking B as reference, the circuit can be redrawn as shown below.
Fig. 9.74
406
Circuits and Networks
If the voltage at o is Vo w.r.t the reference node B.
Then the single-node path equation can be written as
Vo −VAB Vo Vo + VBC
+
+
=0
ZA
ZB
ZC
 1
1
1  VAB VBC
−
Vo 
+
+
+
=0
 Z A Z B ZC  Z A
ZC
 1
1
1  415 0° 415 −120°
=
Vo 
+
+
−
10 0° 8 30° 5 45° 
10 0°
5 45°
Vo (0.1 0° + 0.125 −30° + 0.2 −45°) = 41.5 0°− 83 −165°
Vo (0.404 −30.26°) = 123.55 10°
Vo = 305.8 40.26°
VOA = Vo −VAB = 305.8 40.26°− 415 0° = 268.4 132°
VOC = Vo + VBC = 305.8 40.26° + 415 −120° = 163.83 −80.9°
Line currents/phase currents are given by
IA =
−VOA −268.4 132°
=
= −26.8 132° A
ZA
10 0°
IB =
−VOB −305.8 40.26°
=
= −38.225 10.26° A
ZB
8 30°
IC =
−VOC −163.83 −80.9°
=
= −32.766 −125.9° A
5 45°
ZC
PROBLEM 9.25
A 3-phase, 4-wire, 380 V supply is connected to an unbalanced load having phase impedances of ZR 5 4 1 j3;
ZY 5 4 2 j3; and ZB 5 2 V . Impedance of the neutral wire is Zn 5 (1 1 j2) V. Find the phase currents and
voltages of the load using Millman’s theorem.
Solution
Fig. 9.75
Polyphase Circuits
YR =
1
1
=
= 0.16 − j 0.12 � = 0.2 −36.86°
Z R 4 + j3
YY =
1
1
=
= 0.16 + j 0.12 � = 0.2 36.86°
ZY
4 − j3
YB =
1
1
= = 0.5 � = 0.5 0°
ZB 2
1
= 0.2 − j 0.4 = 0.447 −63.43°
1+ j2
380
Assuming RYB sequence, VR =
0° = 219.4 0°
3
YN =
Similarly,
VY = 219.4 −12°, VB = 219.4 −240°
According to Millman’s theorem, the voltage between two nodes is given by
VN ′ N =
VRYR + VY YY + VBYB
YR + YY + YB + YN
VN ′ N =
( 219.4 0°)(0.2 −36.86°) + ( 219.4 −120°)(0.2 −36.86°) + ( 219.4 −240°)(0.5)
0.2 −36.86° + 0.2 36.86° + 0.5 0° + 0.447 −63.43°
=
43.88 −36.86° + 43.88 −83.14° + 109.7 −240°
1.02 − j 0.4
=
−14.51 + j 25.12
1.02 − j 0.4
=
29 120°
1.09 −21.4°
= 26.6 141.4°
VN ′ N = 26.6 141.4°
= −20.93 + j16.41
The three phase voltages can be found as
VR′ = VR −VN ′ N
= 219.4 − (−20.93 + j16.41)
= 2409 −39°
V ′Y = VY −VN ′ N
= (−110 + j190) − (−20.93 + j16.41)
= 256.3 246.65°
407
408
Circuits and Networks
VB′ = VB −VN ′ N
= (−110 + j190) − (−20.93 + j16.41)
= 195.1117.16°
Phase currents are given by
I R = VR′YY = ( 290.9 −3.9°) (0.2 −36.86°) = 48.18 −40.76°
IY = VY′YY = ( 256
6.3 246.65°) (0.2 36.86°) = 51.26 283.51°
I B = VB′YB = (195.1117.16°)(0.5 0°) = 97.55 117.10°
I N = VN′ YN = ( 26.5 141.9°)(0.447 −63.43°) = 11.89 78.4°
PSpice Problems
PROBLEM 9.1
A 3-phase, D-connected RYB system is shown in
Fig. 9.76 with an effective voltage of 400 V, has a
balanced load with impedances 3 1 j4 V. Using
PSpice, calculate (a) phase currents, (b) line
currents, and (c) power in each phase.
L 5 12.73 mH
R53V
Fig. 9.76 (a)
Fig. 9.76 (b)
* 3 PHASE BALANCED ANALYSIS
VR 1
0
AC 230.94 -30
VY 3
0
AC 230.94 -150
VB 2
0
AC 230.94 90
R1 2
4
3
L1
1
4
12.732 M
R2 2
5
3
L2
5
3
12.732 M
R3 3
6
3
L3
1
6
12.732 M
Polyphase Circuits
409
.AC LIN 1 50 50
.PRINT AC IM(VR) IP(VR) IM(VY) IP(VY) IM(VB) IP(VB)
1 IM(R1) IP(R1) IM(R2) IP(R2) IM(R3) IP(R3)
.PROBE
.END
**** AC ANALYSIS TEMPERATURE 5 27.000 DEG C
****************************************************************
FREQ
IM(VR)
IP(VR)
IM(VY)
IP(VY)
IM(VB)
5.000E101 1.386E102ty 9.687E101 1.386E102 –2.313E101 1.386E102
FREQ
IP(VB)
IM(R1)
IP(R1)
IM(R2)
IP(R2)
5.000E101 –1.431E102 8.000E101 6.687E101 8.000E101 6.871E100
FREQ
IM(R3)
IP(R3)
5.000E101
8.000E101
1.269E102
Result
IR 5 –I(R3) 5 48.03 – J63.97
IY 5 –I(R2) 5 –79.42 – J9.61
IB 5 I(R1) 5 31.38 1 J73.58
I1 5 IR – IB 5 138.55∠–83.09
I2 5 IY – IR 5 138.55∠156.9
I3 5 IB – IY 5 138.55∠36.89
POWER CONSUMED IN EACH PHASE 5 VPH
TOTAL POWER 5 3 3 19.2 K 5 57.6 kW
IPH COS f 5 400 3 80 3 0.6 5 19.2 kW
PROBLEM 9.2
The ac circuit of Fig. 9.77 is supplied from a three-phase balanced supply. Use PSpice to calculate RMS
magnitudes and phase angles of currents, IR, IY , IB, and IN.
Fig. 9.77 (a)
410
Circuits and Networks
9.2
VRN
VYN
VBN
RR
RY
RB
RXL
RYL
RZL
Fig. 9.77 (b)
THREE PHASE CIRCUIT
1
0
AC 120 0
2
0
AC 120 120
3
0
AC 120 -120
1
4
0.5
2
5
0.5
3
6
0.5
4
7
1
5
8
1
6
9
1
R1
7
10
5
R2
8
11
10
R3
9
12
10
C1
10 12
150 UF
L2
11 12
120 MH
VX
12 0
DC 0V
.AC LIN 1 50 100
.PRINT AC IM(RR) IP(RR) IM(RY) IP(RY) IM(RB) IP(RB)
1 VM(7,12) VP(7,12) VM(8,12) VP(8,12) VM(9,12) VP(9,12)
.PRINT AC IM(VX) IP(VX)
.END
**** AC ANALYSIS TEMPERATURE 5 27.000 DEG C
****************************************************************
FREQ
IM(RR)
IP(RR)
IM(RY)
IP(RY)
IM(RB)
5.000E101 5.407E100 7.297E101 3.045E100 4.696E101 1.043E101
FREQ
IP(RB)
VM(7,12)
VP(7,12) VM(8,12) VP(8,12)
5.000E101 –1.200E102 1.179E102 –3.772E100 1.187E102 1.221E102
FREQ
5.000E101
FREQ
5.000E101
VM(9,12)
1.043E102
IM(VX)
2.262E100
VP(9,12)
–1.200E102
IP(VX)
–1.335E102
Answers to Practice Problems
9-4.1 Taking VRN reference
iR 5 25.4 ∠0°; iY 5 25.4 ∠– 146.8°; iB 5 25.4 ∠146.8°
iN 5 17.1 ∠0°; Power 5 17967.7 W
9-4.3
Delta impedances:
Z RY = 3.822 −5.97° V; ZYB = 5.3 −9.15° V and Z BR = 8.546 −110.59° V
Line or phase currents in star load:
63.83 −146.55° A; 169.66 −118.12° A; 117.58 −103.13° A
Phase currents in delta load: 115.12 35.97°; 83.01 −80.85°; 51.486 −320.59°
Power consumed = 78.4 kW
411
Polyphase Circuits
9-4.4
Powers PR = 8007.69 W; PY = 17,923.37 W; PB = 6998 W
Total power = 32.929 kW
9-6.1
iR 5 50.8 ∠0°; iY 5 25.4 ∠–120°; iB 5 16.936 ∠120°
PR 5 12903.2 W; PY 5 6451.6 W; PB 5 4302.4 W
(Taking R-phase voltage reference)
9-6.3
Taking VRY reference
iR 5 11.25 ∠–23.42°; iY 5 18.06 ∠ 218.25°; iB 5 16.12 ∠ 76°
281.32 ∠–23.42°; 180.6 ∠ 218.25°; 241.8 ∠ 76°
9-6.4
VON = 87.43 −142.98° V
9-7.1
3000 W
9-7.3
iR 5 50 ∠– 62°; iY 5 50 ∠– 182°; iB 5 50 ∠ 58°; Power 5 12705 W
9-7.4
173 mF
9-8.4
R = 9.72 V; L = 25.9 mH; P = 23.352 kW; W1=3 kW; W2=8.6 kW
9-9.2
Phase currents : 10 −90°; 15.625 −140°; 12.5 120°
Active powers : 0 W; 3662 W; 3125 W
Reactive powers : 2500 VAR; 1333 VAR; 0 VAR
Apparent power : 7794.5 29° VA
9-11.1 1250 W; 0.693; 2.36 A; 1000 W
Objective-Type Questions
rrr 9.1
The resultant voltage in a closed balanced delta circuit is given by
(a) three times the phase voltage
(c) zero
rrr 9.2
(b)
3 times the phase voltage
Three coils A, B, C, displaced by 120° from each other are mounted on the same axis and rotated in a
uniform magnetic field in clockwise direction. If the instantaneous value of emf in coil A is Emax sin vt, the
instantaneous value of emf in B and C coils will be


2p 
4p 
(a ) Emax sin vt − ; Emax sin vt − 


3 
3 


2p 
4p 
( b) Emax sin vt + ; Emax sin vt +




3 
3 


2p 
4p 
( c) Emax sin vt − ; Emax sin vt +



3 
3 
rrr 9.3
The current in the neutral wire of a balanced three-phase, four-wire star connected load is given by
(a) zero
(b) 3 times the current in each phase
(c) 3 times the current in each phase
rrr 9.4
In a three-phase system, the volt ampere rating is given by
(a) 3VL IL
(b)
3 VL IL
(c) VL IL
412
rrr 9.5
Circuits and Networks
In a three-phase balanced star connected system, the phase relation between the line voltages and their
respective phase voltage is given as under
(a) the line voltages lead their respective phase voltages by 30°
(b) the phase voltages lead their respective line voltage by 30°
(c) the line voltages and their respective phase voltages are in phase
rrr 9.6
In a three-phase balanced delta connected system, the phase relation between the line currents and their
respective phase currents is given by
(a) the line currents lag behind their respective phase currents by 30°
(b) the phase currents lag behind their respective line currents by 30°
(c) the line currents and their respective phase currents are in phase
rrr 9.7
In a three-phase unbalanced, four-wire star-connected system, the current in the neutral wire is given by
(a) zero
(b) three times the current in individual phases
(c) the vector sum of the currents in the three lines
rrr 9.8
In a three-phase unbalanced star-connected system, the vector sum of the currents in the three lines is
(a) zero
(b) not zero
(c) three times the current in the each phase
rrr 9.9
Wattmeter deflection in ac circuit is proportional to the
(a) maximum power in the circuit
(b) instantaneous power in the circuit
(c) average power in the circuit
rrr 9.10
The three-wattmeter method of power measurement can be used to measure power in
(a) balanced circuits
(b) unbalanced circuits
(c) both balanced and unbalanced circuits
rrr 9.11
The two-wattmeter method of power measurement can be used to measure power in
(a) balanced circuits
(b) unbalanced circuits
(c) both balanced and unbalanced circuits
rrr 9.12
In the two-wattmeter method of power measurements, when the pf is 0.5,
(a) the readings of the two wattmeters are equal and positive
(b) the readings of the two wattmeters are equal and opposite
(c) the total power is measured by only one wattmeter
rrr 9.13
The reading of the wattmeter connected to measure the reactive power in a three-phase circuit is given by
zero, the line voltage is 400 V and line current 15 A; then the pf of the circuit is
(a) zero
(b) unity
(c) 0.8
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10
LEARNING OBJECTIVES
10.1
INTRODUCTION
Two circuits are said to be ‘coupled’ when energy transfer takes place from one circuit to the other when
one of the circuits is energised. There are many types of couplings like conductive coupling as shown by the
potential divider in Fig. 10.1 (a), inductive or magnetic coupling as shown by a two-winding transformer
in Fig. 10.1 (b), or conductive and inductive coupling as shown by an auto transformer in Fig. 10.1 (c). A
majority of the electrical circuits in practice are conductively or electromagnetically coupled. Certain coupled
elements are frequently used in network analysis and synthesis. Transformer, transistor, and electronic pots,
Fig. 10.1
414
Circuits and Networks
etc., are some among these circuits. Each of these elements may be represented as a two-port network as
shown in Fig. 10.1 (d).
10.2
CONDUCTIVELY COUPLED CIRCUIT AND MUTUAL IMPEDANCE
A conductively coupled circuit which does not involve magnetic coupling is
LO 1
shown in Fig. 10.2 (a).
In the circuit shown, the impedance Z12 or Z21 common to loops 1 and 2 is
called mutual impedance. It may consists of a pure resistance, a pure inductance,
a pure capacitance, or a combination of any of these elements. Mesh analysis, nodal
analysis or Kirchhoff’s laws can be used to solve these type of circuits as described in Chapter 7.
The general definition of mutual impedance is explained with the help of Fig. 10.2 (b).
(a)
Fig. 10.2
(b)
The network in the box may be of any configuration of circuit elements with two ports having two pairs
of terminals 1-1′ and 2-2′ available for measurement. The mutual impedance between port 1 and 2 can be
measured at 1-1′ or 2-2′. If it is measured at 2-2′. It can be defined as the voltage developed (V2) at 2-2′
per unit current (I1) at port 1-1′. If the box contains linear bilateral elements, then the mutual impedance
measured at 2-2′ is same as the impedance measured at 1-1′ and is defined as the voltage developed (V1) at
1-1′ per unit current (I2) at port 2-2′.
EXAMPLE 10.1
Find the mutual impedance for the circuit shown in Fig. 10.3.
Solution Mutual impedance is given by
V2
V
or 1
I1
I2
V2 =
or
10.3
V
3
I1 or 2 = 1.5 V
I1
2
V1 = 5× I 2 ×
Fig. 10.3
V
3
or 2 = 1.5 V
I2
10
MUTUAL INDUCTANCE
The property of inductance of a coil was introduced in Section 1.6. A voltage
is induced in a coil when there is a time rate of change of current through it.
The inductance parameter L, is defined in terms of the voltage across it and the
di(t )
, where v(t) is the voltage
time rate of change of current through it v (t ) = L
dt
LO 2
415
Coupled Circuits
across the coil, I(t) is the current through the coil and L is the inductance of the coil. Strictly speaking, this
definition is of self-inductance and this is considered as a circuit element with a pair of terminals. Whereas
in a circuit element “mutual inductor” does not exist. Mutual inductance is a property associated with two or
more coils or inductors which are in close proximity and the presence of common magnetic flux which links
the coils. A transformer is such a device whose operation is based on mutual inductance.
Let us consider two coils, L1 and L2, as shown in Fig. 10.4 (a), which are sufficiently close together, so
that the flux produced by i1 in the coil L1 also link the coil L2. We assume that the coils do not move with
respect to one another, and the medium in which the flux is established has a constant permeability. The two
coils may be also arranged on a common magnetic core, as shown in Fig. 10.4 (b). The two coils are said
to be magnetically coupled, but act as a separate circuits. It is possible to relate the voltage induced in one
coil to the time rate of change of current in the other coil. When a voltage v1 is applied across L1, a current
i1 will start flowing in this coil, and produce a flux f. This flux also links the coil L2. If i1 were to change
with respect to time, the flux ‘f’ would also change with respect to time. The time-varying flux surrounding
the second coil, L2 induces an emf, or voltage, across the terminals of L2; this voltage is proportional to
the time rate of change of current flowing through the first coil L1. The two coils, or circuits, are said to be
inductively coupled, because of this property they are called coupled elements or coupled circuits and the
di1 (t )
induced voltage, or emf is called the voltage/emf of mutual induction and is given by v2 (t ) = M1 dt
volts, where v2 is the voltage induced in coil L2 and M1 is the coefficient of proportionality, and is called the
coefficient of mutual inductance, or simple mutual inductance.
If current i2 is made to pass through coil L2 as shown in Fig. 10.4 (c) with coil L1 open, a change of i2
would cause a voltage v1 in coil L1, given by v (t ) = M di2 (t ) .
1
2
dt
In the above equation, another coefficient of proportionality M2 is involved. Though it appears that two
mutual inductances are involved in determining the mutually induced voltages in the two coils, it can be
shown from energy considerations that the two coefficients are equal and, therefore, need not be represented
by two different letters. Thus, M1 5 M2 5 M.
di1 (t )
volts
dt
di (t )
v1 (t ) = M 2
volts
dt
∴ v2 ( t ) = M
In general, in a pair of linear time invariant coupled coils or inductors, a non-zero current in each of the two
coils produces a mutual voltage in each coil due to the flow of current in the other coil. This mutual voltage is
present independently of, and in addition to, the voltage due to self induction. Mutual inductance is also measured
(c)
Fig. 10.4
416
Circuits and Networks
di
may be either positive or negative, depending
dt
on the physical construction of the coil and reference directions. To determine the polarity of the mutually induced
voltage (i.e. the sign to be used for the mutual inductance), the dot convention is used.
in henries and is positive, but the mutually induced voltage, M
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 2*
rrr 10-2.1
rrr 10-2.2
rrr 10-2.3
rrr 10-2.4
rrr 10-2.5
Two inductively coupled coils have self-inductances L1540 mH and L25150 mH. If the
coefficient of coupling is 0.7, (a) find the value of mutual inductance between the coils, and
(b) the maximum possible mutual inductance.
Two coils connected in series have an equivalent inductance of 0.8 H when connected in aiding,
and an equivalent inductance of 0.5 H when the connection is opposing. Calculate the mutual
inductance of the coils.
Calculate the effective inductance of the circuit shown in Fig. Q.3 across XY.
Calculate the effective inductance of the circuit shown in Fig. Q.4.
Find the equivalent inductance between the terminals across ab for the coupled circuit shown,
M 5 0.5 H. All the coils are coupled. (Fig. Q.5)
a
1H
1H
b
1H
Fig. Q.5
Fig. Q.4
Fig. Q.3
Frequently Asked Questions linked to LO 2*
rrr
rrr
j6 W
j5 W
j10 W
B
A
k = 0.5
J 10 W
C
j 5W
3W
3W
50Ð0ºV
I1
I2 5 W
5W
50Ð0º
I2
I1
–J 4 W
–j4 W
F
Fig. Q.1
rrr 10-2.3
E
D
Fig. Q.2
Write the expression which relates the self and mutual inductance.
[AU May/June 2014]
Define mutual inductance.
[AU Nov./Dec. 2012]
rrr 10-2.5 Two coupled coils have self-inductances of L1 = 100 mH and L2 = 400 mH. The coupling
rrr 10-2.4
417
Coupled Circuits
coefficient is 0.8. Find M. If N1 is 1000 turns, what is the values of N2? If a current i1 = 2 sin (500t)
A through the coil 1, find the flux f1 and the mutually induced voltage V2M. [AU Nov./Dec. 2012]
rrr 10-2.6
Two inductively coupled coil have self-inductances L1 =
50 mH and L2 = 200 mH. If the coefficient of coupling is
0.5, compute the value of mutual inductance between the
coils.
[AU April/May 2011]
rrr 10-2.7 For the network shown in Fig. Q.7, find the voltage across
the load resistance RL.
[JNTU Nov. 2012]
10.4
k = 0.5
j5 W
V
j2 W
j3W
RL = 5 W
Fig. Q.7
DOT CONVENTION
Dot convention is used to establish the choice of correct sign for the mutually
LO 3
induced voltages in coupled circuits.
Circular dot marks and/or special symbols are placed at one end of each
of two coils which are mutually coupled to simplify the diagrammatic
representation of the windings around its core.
Let us consider Fig. 10.5, which shows a pair of linear, time-invariant, coupled
inductors with self- inductances L1 and L2 and a mutual inductance M. If these inductors form a portion of a
network, currents i1 and i2 are shown, each arbitrarily assumed entering at the dotted terminals, and voltages v1
and v2 are developed across the inductors. The voltage across L1 is, thus composed of two parts and is given by
v1 (t ) = L1
di1 (t )
di (t )
±M 2
dt
dt
The first term on the RHS of the above equation is the self induced voltage
due to i1, and the second term represents the mutually induced voltage due to
i2.
Similarly,
v2 (t ) = L2
di2 (t )
di (t )
±M 1
dt
dt
Although the self-induced voltages are designated with positive sign,
mutually
induced voltages can be either positive or negative depending on the
Fig. 10.5
direction of the winding of the coil and can be decided by the presence of the
dots placed at one end of each of the two coils. The convention is as follows.
If two terminals belonging to different coils in a coupled circuit are marked identically with dots then for
the same direction of current relative to like terminals, the magnetic flux of self- and mutual induction in each
coil add together. The physical basis of the dot convention can be verified by examining Fig. 10.6. Two coils
ab and cd are shown wound on a common iron core.
It is evident from Fig. 10.6 that the direction of the winding of the coil ab is clock - wise around the core as
viewed at X, and that of cd is anti-clockwise as viewed at Y. Let the direction of the current i1 in the first coil
be from a to b, and increasing with time. The flux produced by i1 in the core has a direction which may be
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
418
Circuits and Networks
Fig. 10.6
Fig. 10.7
found by right hand rule, and which is downwards in the
left limb of the core. The flux also increases with time in
the direction shown at X. Now suppose that the current i2
in the second coil is from c to d, and increasing with time.
The application of the right hand rule indicates that the
flux produced by i2 in the core has an upward direction in
the right limb of the core. The flux also increases with time
in the direction shown at Y. The assumed currents i1 and
i2 produce flux in the core that are additive. The terminals
a and c of the two coils attain similar polarities simultaneously. The two
simultaneously positive terminals are identified by two dots by the side of
the terminals as shown in Fig. 10.7.
The other possible location of the dots is the other ends of the coil to
get additive fluxes in the core, i.e. at b and d. It can be concluded that the
mutually induced voltage is positive when currents i1 and i2 both enter (or
leave) the windings by the dotted terminals. If the current in one winding
enters at the dot-marked terminals and the current in the other winding
leaves at the dot-marked terminal, the voltages due to self and mutual
induction in any coil have opposite signs.
EXAMPLE 10.2
Using dot convention, write voltage equations for the coils shown in
Fig. 10.8.
Solution Since the currents are entering at the dot-marked terminals,
the mutually induced voltages or the sign of the mutual inductance is
positive; using the sign convention for the self inductance, the equations
for the voltages are
di
di
v1 = L1 1 + M 2
dt
dt
di2
di
v2 = L2
+M 1
dt
dt
EXAMPLE 10.3
Write the equation for voltage v0 for the circuits shown in Fig. 10.9.
Fig. 10.8
419
Coupled Circuits
Fig. 10.9
Solution
v0 is assumed positive with respect to the terminal C and the equation is given by
di
di
di
(a) v0 = M
(b) v0 = − M
(c) v0 = − M
(d)
dt
dt
dt
v0 = M
di
dt
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 3
r 10-3.1
Using the dot convention, write the voltage equations for the coils shown in Fig. Q.3.
Fig. Q.3
Frequently Asked Questions linked to LO 3
r 10-3.1
What is dot convention in coupled circuit?
[BPUT 2007]
rr 10-3.2 Explain the dot convention rule, for the magnetically coupled network using network shown in
Fig.Q.2. Also formulate KVL equations.
[GTU Dec. 2012]
RL
I2
I1 r1
V1
L1
L2
Fig. Q.2
10.5
COEFFICIENT OF COUPLING
The amount of coupling between the inductively coupled coils is expressed in
LO 4
terms of the coefficient of coupling, which is defined as K = M / L1 L2
where
M 5 mutual inductance between the coils,
L1 5 self-inductance of the first coil, and
L2 5 self-inductance of the second coil.
Coefficient of coupling is always less than unity, and has a maximum value of 1 (or 100%). This case, for
420
Circuits and Networks
which K 5 1, is called perfect coupling, when the entire flux of one coil
links the other. The greater the coefficient of coupling between the two coils,
the greater the mutual inductance between them, and vice versa. It can be
expressed as the fraction of the magnetic flux produced by the current in one
coil that links the other coil.
For a pair of mutually coupled circuits shown in Fig. 10.10, let us assume
initially that i1, i2 are zero at t 5 0.
di (t )
di (t )
Then v1 (t ) = L1 1 + M 2
dt
dt
and
v2 (t ) = L2
Fig. 10.10
di2 (t )
di (t )
+M 1
dt
dt
Initial energy in the coupled circuit at t 5 0 is also zero. The net energy input to the system shown in Fig.
10.10 at time t is given by
t
W (t ) = ∫ [ v1 (t ) i1 (t ) + v2 (t ) i2 (t ) ] dt
0
Substituting the values of v1(t) and v2(t) in the above equation yields
t

di (t )
di (t )
W (t ) = ∫  L1i1 (t ) 1 + L2 i2 (t ) 2

dt
dt
0 
+ M (i1 (t ))
di2 (t )
di (t ) 
+ i2 (t ) 1  dt
dt
dt 
from which, we get
1
1
2
2
W (t ) = L1 [i1 (t ) ] + L2 [i2 (t ) ] + M [i1 (t ) i2 (t ) ]
2
2
If one current enters a dot-marked terminal while the other leaves a dot-marked terminal, the above
equation becomes
1
1
2
2
L1 [i1 (t ) ] + L2 [i2 (t ) ] − M [i1 (t ) i2 (t ) ]
2
2
According to the definition of passivity, the net electrical energy input to the system is non-negative. W(t)
represents the energy stored within a passive network, it cannot be negative.
W (t ) =
∴ W (t) $ 0 for all values of i1, i2; L1, L2 or M
The statement can be proved in the following way. If i1 and i2 are both positive or negative, W(t) is positive.
The other condition where the energy equation could be negative is
W (t ) =
1
1
2
2
L1 [i1 (t ) ] + L2 [i2 (t ) ] − M [i1 (t ) i2 (t ) ]
2
2
The above equation can be rearranged as
2
1 
M 
1
M 2  2
W (t ) =  L1i1 −
i2  +  L2 −
 i2
2 
2 
L1 
L1 
421
Coupled Circuits
The first term in the parenthesis of the right side of the above equation is positive for all values of i1 and
i2, and, thus, the last term cannot be negative; hence,
M2
≥0
L1
L1 L2 − M 2
≥0
L1
L2 −
L1L2 – M 2 ≥ 0
L1 L2 ≥ M
M ≤ L1 L2
Obviously, the maximum value of the mutual inductance is
coupling for the coupled circuit as
L1 L2 . Thus, we define the coefficient of
M
K=
L1 L2
The coefficient, K, is a non-negative number and is independent of the reference directions of the currents
in the coils. If the two coils are a great distance apart in space, the mutual inductance is very small, and K is
also very small. For iron-core coupled circuits, the value of K may be as high as 0.99, for air-core coupled
circuits, K varies between 0.4 to 0.8.
EXAMPLE 10.4
Two inductively coupled coils have self inductances L1 5 50 mH and L2 5 200 mH. If the coefficient of coupling
is 0.5 (a), find the value of mutual inductance between the coils, and (b) what is the maximum possible mutual
inductance?
M = K L1 L2
Solution (a)
= 0.5 50 ×10−3 × 200 ×10−3 = 50 ×10−3 H
Maximum value of the inductance when K 5 1,
(b)
M = L1 L2 = 100 mH
Frequently Asked Questions linked to LO 4
r
r
N
N
I1
M
rr
f
N
f
f12
N
f11
i
K,
L
L
M,
N1
N2
Fig. Q.3
422
10.6
Circuits and Networks
IDEAL TRANSFORMER
Transfer of energy from one circuit to another circuit through mutual induction
LO 5
is widely utilised in power systems. This purpose is served by transformers.
Most often, they transform energy at one voltage (or current) into energy at
some other voltage (or current).
A transformer is a static piece of apparatus, having two or more windings or coils arranged on a common
magnetic core. The transformer winding to which the supply source is connected is called the primary, while
the winding connected to load is called the secondary. Accordingly, the voltage across the primary is called
the primary voltage, and that across the secondary, the secondary voltage. Correspondingly, i1 and i2 are the
currents in the primary and secondary windings. One such transformer is shown in Fig. 10.11 (a). In circuit
diagrams, ideal transformers are represented by Fig. 10.11 (b).
Fig. 10.11
The vertical lines between the coils represent the iron core; the currents are assumed such that the mutual
inductance is positive. An ideal transformer is characterised by assuming (i) zero power dissipation in the
primary and secondary windings, i.e. resistances in the coils are assumed to be zero, (ii) the self inductances
of the primary and secondary are extremely large in comparison with the load impedance, and (iii) the
coefficient of coupling is equal to unity, i.e. the coils are tightly coupled without having any leakage flux. If
the flux produced by the current flowing in a coil links all the turns, the self inductance of either the primary
or secondary coil is proportional to the square of the number of turns of the coil. This can be verified from
the following results.
The magnitude of the self induced emf is given by
di
v=L
dt
If the flux linkages of the coil with N turns and current are known, then the self- induced emf can be
expressed as
df
v=N
dt
L
df
di
=N
dt
dt
L=N
f=
df
di
Ni
reluctance
Coupled Circuits
∴
L=N
L=
423

d 
Ni


di  reluctance 
N2
reluctance
L a N2
From the above relation, it follows that
L2
N2
= 22 = a 2
L1
N1
where a 5 N2/N1 is called the turns ratio of the transformer. The turns ratio, a, can also be expressed in terms of
primary and secondary voltages. If the magnetic permeability of the core is infinitely large then the flux would be
confined to the core. If f is the flux through a single turn coil on the core and N1, N2 are the number of turns of the
primary and secondary, respectively, then the total flux through windings 1 and 2, respectively, are
f1 5 N1 f; f2 5 N2 f
d f1
df2
, and v2 =
dt
dt
df
N2
dt = N 2
so that v2 =
df
v1
N1
N1
dt
Figure 10.12 shows an ideal transformer to which the secondary is
N
Fig. 10.12
connected to a load impedance ZL. The turns ratio 2 = a.
N1
The ideal transformer is a very useful model for circuit calculations, because with few additional elements
like R, L, and C, the actual behaviour of the physical transformer can be accurately represented. Let us
analyse this transformer with sinusoidal excitations. When the excitations are sinusoidal voltages or currents,
the steady state response will also be sinusoidal. We can use phasors for representing these voltages and
currents. The input impedance of the transformer can be determined by writing mesh equations for the circuit
shown in Fig. 10.12.
Also, we have
V1 5 jvL1I1 – jvMI2
0 5 – jvMI1 1 (ZL 1 jvL2)I2
v1 =
(10.1)
(10.2)
where V1, V2 are the voltage phasors, and I1, I2 are the current phasors in the two windings. jvL1 and jvL2 are
the impedances of the self inductances and jvM is the impedance of the mutual inductance, v is the angular
frequency.
From Eq. (10.2),
I2 =
jvMI1
( Z L + jvL2 )
Substituting in Eq. (10.1), we have
V1 = I1 jvL1 +
I1v2 M 2
Z L + jvL2
424
Circuits and Networks
The input impedance Zin =
∴
Zin = jvL1 +
V1
I1
v2 M 2
( Z L + jvL2 )
When the coefficient of coupling is assumed to be equal to unity,
M = L1 L2
∴
Zin = jvL1 +
v2 L1 L2
( Z L + jvL2 )
We have already established that
L2
= a2
L1
where a is the turns ratio N2 / N1.
∴
Zin = jvL1 +
v2 L12 a 2
( Z L + jvL2 )
Further simplication leads to
Zin =
( Z L + jvL2 ) jvL1 + v2 L12 a 2
( Z L + jvL2 )
Zin =
jvL1Z L
( Z L + jvL2 )
As L2 is assumed to be infinitely large compared to ZL,
Zin =
jvL1Z L
jva 2 L1
=
Fig. 10.13
 N1 2
  Z
=
 L
a 2  N 2 
ZL
The above result has an interesting interpretation, that is the
ideal transformers change the impedance of a load, and can be used
to match circuits with different impedances in order to achieve
maximum power transfer. For example, the input impedance of a
loudspeaker is usually very small, say 3 V to 12 V, for connecting
directly to an amplifier. The transformer with proper turns ratio
can be placed between the output of the amplifier and the input of
the loudspeaker to match the impedances as shown in Fig. 10.13.
EXAMPLE 10.5
An ideal transformer has N1 5 10 turns, and N2 5 100 turns. What is the value of the impedance referred to
as the primary, if a 1000 V resistor is placed across the secondary?
Solution
100
= 10
10
Z
1000
Zin = 2L =
= 10 V
100
a
The turns ratio a =
425
Coupled Circuits
The primary and secondary currents can also be expressed in terms of turns ratio. From Eq. (10.2), we
have
I1 jvM 5 I2(ZL 1 jvL2)
I1
Z + jvL2
= L
I2
j vM
When L2 is very large compared to ZL,
I1
jvL2
L
=
= 2
I2
j vM
M
Substituting the value of M = L1 L2 in the above equation,
I1
=
I2
L2
L1 L2
I1
=
I2
L2
N
=a= 2
L1
N1
=
L2
L1
I1
L
= 2
I2
M
EXAMPLE 10.6
An amplifier with an output impedance of 1936 V is to feed a loudspeaker with an impedance of 4 V.
(a) Calculate the desired turns ratio for an ideal transformer to connect the two systems.
(b) An rms current of 20 mA at 500 Hz is flowing in the primary. Calculate the rms value of current in the
secondary at 500 Hz.
(c) What is the power delivered to the load?
Load impedance
Solution (a) To have maximum power transfer, the output impedance of the amplifier =
a2
4
∴ 1936 = 2
a
∴
a=
4
1
=
1936 22
N2
1
=
N1 22
(b) I1 5 20 mA
I1
=a
I2
RMS value of the current in the secondary winding
We have
=
I1 20 ×10−3
= 0.44 A
=
a
1 / 22
(c) The power delivered to the load (speaker)
5 (0.44)2 3 4 5 0.774 W
426
Circuits and Networks
The impedance changing properties of an ideal transformer may be utilised to simplify circuits. Using this
property, we can transfer all the parameters of the primary side of the transformer to the secondary side, and
vice versa. Consider the coupled circuit shown in Fig. 10.14 (a).
To transfer the secondary side load and voltage to the primary side, the secondary voltage is to be divided
by the ratio, a, and the load impedance is to be divided by a2. The simplified equivalent circuits is shown in
Fig. 10.14 (b).
Fig. 10.14
EXAMPLE 10.7
For the circuit shown in Fig. 10.15 with turns ratio, a 5 5, draw the equivalent circuit referring (a) to
primary, and (b) secondary. Take source resistance as 10 V.
Fig. 10.15
Solution (a)
Equivalent circuit referred to primary is as shown in Fig. 10.16 (a).
Fig. 10.16 (a)
427
Coupled Circuits
(b) Equivalent circuit referred to secondary is as shown in Fig. 10.16 (b).
Fig. 10.16 (b)
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 5
In Fig. Q.1, L1 5 2 H; L2 5 6 H; K 5 0.5; i1 5 4 sin (40t 2 30°) A; i2 5 2 sin (40t 2 30°) A.
Find the values of (a) v1, and (b) v2.
rrr 10-5.2 For the circuit shown in Fig. Q.2, write the mesh equations.
rrr 10-5.1
Fig. Q.1
rrr 10-5.3
Fig. Q.2
Write the mesh equations for the network shown in Fig. Q.3.
Fig. Q.3
Frequently Asked Questions linked to LO 5
rrr 10-5.1
A voltage of 100 V at a frequency of 106/2p Hz is applied to the primary of coupled circuit.
Calculate primary and secondary currents.
[RTU Feb. 2011]
428
Circuits and Networks
K = 0.53
r1 = 15W
r2 = 8 W
C2 = 0.01mf
L1 = 200 mH
100Ð0º
C1 = 5000mf
Fig. Q. 1
rrr 10-5.2
OR
In the figure, two coils A (R1 = 5 W, L1 = 0.01 H) and B (R2 = 100 W, L2 = 5 H) have coefficient of
coupling 0.8. Calculating the percentage change in effective resistance of the coil A at a frequency of
50 Hz when resistance connected across terminals of the coil B becomes 0 ohms. [RTU Feb. 2011]
R1 = 5W
V1
I1
M
L1
R2 = 100 W
L2
I2
R = 200W
Fig. Q.2
10.7
ANALYSIS OF MULTI-WINDING COUPLED CIRCUITS
Inductively coupled multi-mesh circuits can be analysed using Kirchhoff’s
LO 6
laws and by loop current methods. Consider Fig. 10.17, where three coils are
inductively coupled. For such a system of inductors, we can define a inductance
matrix L as
L
L12 L13 
 11

L =  L21 L22 L23 
L

 31 L32 L33 
where L11, L22 and L33 are self-inductances of the coupled circuits, and L12 5 L21; L23 5 L32 and L13 5 L31
are mutual inductances. More precisely, L12 is the mutual inductance between coils 1 and 2, L13 is the mutual
inductance between coils 1 and 3, and L23 is the mutual inductance between coils 2 and 3. The inductance
matrix has its order equal to the number of inductors and is symmetric. In terms of voltages across the coils,
we have a voltage vector related to i by
 di 
[v] = [ L]  
 dt 
where v and i are the vectors of the branch voltages and currents, respectively. Thus, the branch volt-ampere
relationships of the three inductors are given by
 v1   L11 L12 L13   di1 /dt 
  
 v2  =  L21 L22 L23   di2 /dt 
  
 v3   L31 L32 L33   di3 /dt 
Coupled Circuits
429
Fig. 10.17
Using KVL and KCL, the effective inductances can be calculated. The polarity for the inductances can be
determined by using passivity criteria, whereas the signs of the mutual inductances can be determined by
using the dot convention.
EXAMPLE 10.8
For the circuit shown in Fig. 10.18, write the inductance matrix.
Fig. 10.18
Let L1, L2, and L3 be the self-inductances, and L12 5 L21, L23 5 L32 and L13 5 L31 be the mutual
inductances between coils, 1, 2, 2, 3 and 1, 3, respectively.
L12 5 L21 is positive, as both the currents are entering at dot marked terminals, whereas L13 5 L31 and
L23 5 L32 are negative.
Solution
430
Circuits and Networks
 L1
L12


∴ The inductance matrix is L =  L21
L2
−L
−
L32
 31
−L13 

−L23 
L3 
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 6
rrr 10-6.1
For the circuit shown in Fig. Q.1, write the inductance matrix.
Fig. Q.1
10.8
SERIES CONNECTION OF COUPLED INDUCTORS
Let there be two inductors connected in series, with self-inductances L1 and L2 and
LO 7
mutual inductance of M. Two kinds of series connections are possible; series aiding
as in Fig. 10.19 (a), and series opposition as in Fig. 10.19 (b).
In the case of series-aiding
connection, the currents in both inductors at any instant
of time are in the same direction relative to like terminals
as shown in Fig. 10.19 (a). For this reason, the magnetic
fluxes of self-induction and of mutual induction linking
with each element add together.
In the case of series-opposition connection, the
currents in the two inductors at any instant of time
are in opposite direction relative to like terminals as
shown in Fig. 10.19 (b). The inductance of an element
is given by L 5 f/i, where f is the flux produced by
the inductor.
Fig. 10.19
∴
f 5 Li
For the series-aiding circuit, if f1 and f2 are the flux produced by the coils 1 and 2, respectively, then the
total flux
f 5 f1 1 f2
431
Coupled Circuits
where f1 5 L1i1 1 Mi2
f2 5 L2i2 1 Mi1
∴ f 5 Li 5 L1i1 1 Mi2 1 L2i2 1 Mi1
Since i1 5 i2 5 i
L 5 L1 1 L2 1 2M
Similarly, for the series opposition,
f 5 f1 1 f2
where f1 5 L1i1 – Mi2
f2 5 L2i2 – Mi1
f 5 Li 5 L1i1 – Mi2 1 L2i2 – Mi1
Since i1 5 i2 5 i
L 5 L1 1 L2 – 2M
In general, the inductance of two inductively coupled elements in series is given by L 5 L1 1 L2 6 2M.
Positive sign is applied to the series aiding connection, and negative sign to the series opposition connection.
EXAMPLE 10.9
Two coils connected in series have an equivalent inductance of 0.4 H when connected in aiding, and an
equivalent inductance 0.2 H when the connection is opposing. Calculate the mutual inductance of the coils.
Solution When the coils are arranged in aiding connection, the inductance of the combination is
L1 1 L2 1 2M 5 0.4; and for opposing connection, it is L1 1 L2 – 2M 5 0.2. Solving the two equations,
we get
4M 5 0.2 H
M 5 0.05 H
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 7
rrr 10-7.1
The inductance matrix for the circuit of a three series connected coupled coils is given below. Find the
inductances and indicate the dots for the coils.
 8 −2 1 


L = −2 4 −6
 1 −6 6 


Frequently Asked Questions linked to LO 7
rrr 10-7.1
What is the expression for the total induce. lance of the three series connected coupled coils shown
in Fig. Q.1.
[BPUT 2007]
M12
M12
L
L1
M13
L2
Fig. Q.1
L3
432
10.9
Circuits and Networks
PARALLEL CONNECTION OF COUPLED COILS
Consider two inductors with self-inductances L1 and L2 connected parallel which
are mutually coupled with mutual inductance M as shown in Fig. 10.20.
LO 8
Fig. 10.20
Let us consider Fig. 10.20 (a) where the self-induced emf in each coil assists the mutually induced emf as
shown by the dot convention.
i 5 i1 1 i2
di di1 di2
=
+
dt
dt
dt
The voltage across the parallel branch is given by
di
di
di
di
v = L1 1 + M 2 or L2 2 + M 1
dt
dt
dt
dt
Also,
L1
(10.3)
di1
di
di
di
+ M 2 = L2 2 + M 1
dt
dt
dt
dt
di1
di
( L1 − M ) = 2 ( L2 − M )
dt
dt
∴
di1 di2 ( L2 − M )
=
dt
dt ( L1 − M )
(10.4)
Substituting Eq. (10.4) in Eq. (10.3), we get
 ( L2 − M ) 


 L − M + 1
 1

If Leq is the equivalent inductance of the parallel circuit in Fig. 10.20 (a) then v is given by
di
v = Leq
dt
di di2 ( L2 − M ) di2
+
=
dt
dt ( L1 − M ) dt
Leq
di
di
di
= L1 1 + M 2
dt
dt
dt
1
di
=
dt Leq
 di1
di 
 L1
+M 2
 dt
dt 
(10.5)
433
Coupled Circuits
Substituting Eq. (10.4) in the above equation, we get
di2 
di
1  di2 ( L2 − M )
=
+
L
M
1
dt Leq  dt ( L2 − M )
dt 
 di
1  ( L2 − M )
 2
(10.6)
=
+
L
M
1
 dt
Leq  ( L1 − M )

Equating Eq. (10.6) and Eq. (10.5), we get

L2 − M
1   L2 − M 
 + M 
L1 
+1 =


L2 − M
Leq   L1 − M 

Rearranging and simplifying the above equation results in
L L −M2
Leq = 1 2
L1 + L2 − 2 M
If the voltage induced due to mutual inductance opposes the self-induced emf in each coil as shown by the dot
convention in Fig. 10.20 (b), the equivalent inductance is given by
Leq =
L1 L2 − M 2
L1 + L2 + 2 M
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 8
rrr 10-8.1
rrr 10-8.2
Find the source voltage if the voltage across 100 ohms is 50 V for the network in Fig. Q.1.
Using PSpice, find V0 in the circuit shown in Fig. Q.2.
Fig. Q.1
Fig. Q.2
Frequently Asked Questions linked to LO 8
rrr 10-8.1
A coil having an inductance of 100 mH is magnetically
coupled to another coil having an inductance of 900
mH. The coefficient of coupled between the coil is
0.45. Calculate the equivalent inductance if the two coil
are connected in (a) series opposing, and (b) parallel
opposing.
[AU May/June 2014]
rrr 10-8.2 Find the voltage drop across the resistance “r” in the
network shown in Fig. Q.2.
[BPUT 2007]
rrr 10-8.3 Two coils in differential connection have self-inductances
of 2 mH and 4 mH and a mutual inductance of 0.15 mH.
K = 0.6
16 W
12 W
r = 10 W
–15 W
+
12Ð0º
–
Fig. Q.2
434
Circuits and Networks
What is the equivalent inductance?
L
rrr
[BPUT 2008]
L
M
10.10
TUNED CIRCUITS
Tuned circuits are, in general, single-tuned and double-tuned. Double-tuned circuits
are used in radio receivers to produce uniform response to modulated signals over
a specified bandwidth; double-tuned circuits are very useful in a communication
system.
LO 9
10.10.1 Single Tuned Circuit
Consider the circuit in Fig. 10.21. A tank circuit (i.e. a parallel resonant circuit) on the secondary side is
inductively coupled to the coil 1 which is excited by a source, vi. Let Rs be the source resistance and R1, R2 be
the resistances of coils 1 and 2, respectively. Also let L1, L2 be the
self inductances of the coils, 1 and 2, respectively.
Let Rs 1 R1 1 jvL1 5 Rs
with the assumption that Rs >> R1 >> jvL1 The mesh equations for
the circuit shown in Fig. 10.21 are
i1Rs – jvMi2 5 vi

j 
− jvMi1 +  R2 + jvL2 −
 i2 = 0

vC 
Fig. 10.21
i2 =
Rs
vi
− j vM
0
i2 =
(− jvM )
Rs

j 
(− jvM )  R2 + jvL2 − 

vC 
jvi vM

j 
2
2
Rs  R2 + jvL2 −
+ v M

vC 
The output voltage
vo = i2 ⋅
vo =
1
j vC
jvi vM


 
1 
2
2
j vC 
 + v M 
 Rs  R2 +  jvL2 −




vC 

 
The voltage transfer function, or voltage amplification, is given by
vo
= A=
vi
 

C
 Rs  R2 +


 
M


1 
2
2
j vL2 −
 + v M 



vC 
Coupled Circuits
435
When the secondary side is tuned, i.e. when the value of the frequency v is such that vL2 5 1/vC, or at
resonance frequency vr, the amplification is given by
v
M
A= o =

vi
C  Rs R2 + vr2 M 2 


The current i2 at resonance i2 =
jvi vr M
Rs R2 + v2r M 2
Thus, it can be observed that the output voltage, current, and amplification depend on the mutual
inductance M at resonance frequency, when M = K L1 L2 . The maximum output voltage or the maximum
amplification depends on M. To get the condition for maximum output voltage, make dvo/dM 5 0.


dvo
vi M

d 
=


2
2


d M d M  C Rs R2 + vr M 
 
 
5 1 2 2M 2 v2r [RsR2 1 v2rM 2]21 5 0
From which, Rs R2 5 v2r M 2
or
M=
Rs R2
vr
From the above value of M, we can calculate the maximum output voltage. Thus,
vi
voM =
,
2vr C Rs R2
or the maximum amplification is given by
Am =
1
2vr C Rs R2
and i2 =
jvi
2 Rs R2
The variation of the amplification factor or output voltage with the coefficient of coupling is shown in Fig.
10.22.
Fig. 10.22
436
Circuits and Networks
EXAMPLE 10.10
Consider the single-tuned circuit shown in Fig. 10.23 and determine (a) the resonant frequency, (b) the output
voltage at resonance, and (c) the maximum output voltage. Assume Rs >> vr L1, and K 5 0.9.
Fig. 10.23
Solution
M = K L1 L2
= 0.9 1×10−6 ×100 ×10−6
= 9 mH
(a) Resonance frequency
1
1
vr =
=
−6
L2C
100 ×10 × 0.1×10−6
=
or
he value of
106
rad/sec.
10
f r = 50.292 kHz
vr L1 =
106
10
1×10−6 = 0.316
Thus, the assumption that vr L1 < Rs is justified.
(b) Output voltage
vo =
Mvi
C  Rs R2 + vr2 M 


9×10−6 ×15
=


 106 2
−6 
 × 9×10−6 
0.1×10 10 ×10 + 
 10 




= 1.5 mV
(c) Maximum value of output voltage
vi
voM =
2vr C Rs R2
Coupled Circuits
15
=
2×
voM
437
106
10
= 23.7 V
× 0.1×10−6 100
10.10.2 Double-Tuned Coupled Circuits
Figure 10.24 shows a double-tuned transformer circuit involving two series resonant circuits.
Fig. 10.24
For the circuit shown in the figure, a special case where the primary and secondary resonate at the same
frequency vr, is considered here,
i.e v2r =
1
1
=
L1C1 L2C2
The two mesh equations for the circuit are

j 
 − i2 jvM
vin = i1  Rs + R1 + jvL1 −

vC1 

j 

0 = − jvMi1 + i2  R2 + jvL2 −

vC2 
from which,
i2 =
Also, vr =

( R + R ) +
1
 s

1
L1C1
=
vin jvM
 
  R2 +
 
1  

1
j vL1 −

vC
1
L2C2
at resonance
vin M

C2 ( Rs + R1 ) R2 + v2r M 2 


or vo = Avin
vo =

1 
 + v2 M 2
j vL2 −

C2 
vC
438
Circuits and Networks
where A is the amplification factor given by
M
A=
C2 ( R1 + Rs ) R2 + v2r M 2 


The maximum amplification or the maximum output voltage can be obtained by taking the first derivative
of vo with respect to M, and equating it to zero.
dvo
dA
∴
= 0, or
=0
dM
dM
dA
= ( R1 + Rs ) R2 + v2r M 2 − 2 M 2 vr2 = 0
dM
v2r M 2 5 R2(R1 1 Rs)
Mc =
R2 ( R1 + Rs )
vr
where Mc is the critical value of mutual inductance. Substituting the value of Mc in the equation of vo, we
obtain the maximum output voltage as
v
vin
vo = 2 in
=
2vr C2 M c 2vr C2 R2 ( R1 + Rs )
vin
vin
i2 =
=
and
2vr M c 2 R2 ( R1 + Rs )
By definition, M = K L1 L2 , the coefficient of coupling, K at M 5 Mc is called the critical coefficient of
coupling, and is given by K c = M c / L2 L1 .
The critical coupling causes the secondary current to have the maximum possible value. At resonance, the
maximum value of amplification is obtained by changing M, or by changing the coupling coefficient for a given
value of L1 and L2. The variation of output voltage with frequency for different coupling coefficients is shown
in Fig. 10.25.
Fig. 10.25
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 9
rrr 10-9.1
For the circuit shown in Fig. Q.1, find the ratio of output voltage to the input voltage.
439
Coupled Circuits
rrr10-9.2
Using PSpice, find I1 and I2 in the circuit shown in Fig. Q.2. Also calculate the power absorbed by the
4 V resistor.
Fig. Q.1
rrr 10-9.3
Fig. Q.2
Using PSpice, find the Thevenin equivalent circuit for the circuit shown in Fig. Q.3 at a2b.
Fig. Q.3
rrr10-9.4
rrr 10-9.5
Solve for the currents I1 and I2 in the circuit shown. Also find the ratio of V2 .
V1
For the coupled circuit shown in Fig. Q.5. Find Leq across xy.
1Ω
+
V1
10 0°
L1 = 1H
L2 = 4H
–
ω = 2r/sec
+
2Ω V2
–
Fig. Q.5
Frequently Asked Questions linked to LO 9
rrr 10-9.1
Give the applications of tuned circuits.
For the circuit shown in Fig. Q.2 determine the voltage
ratio V1/V2. Which will make the current I1 equal to zero.
[AU May/June 2014]
rrr 10-9.3 Derive the expressions for maximum output voltage and
maximum amplification of a single-tuned circuit.
[AU April/May 2011]
rrr 10-9.4 Explain in detail the following:
[RGTU June 2014]
(a) Double-tuned circuit
(b) Single-tuned air-core transformer
[AU May/June 2013]
rrr 10-9.2
5W
V1
j2W
2W
j4W
j8W
I1
I2
Fig. Q.2
V2
440
Circuits and Networks
10.11
ANALYSIS OF MAGNETIC CIRCUITS
The presence of charges in space or in a medium creates an electric field; similarly,
LO 10
the flow of current in a conductor sets up a magnetic field. Electric field is represented
by electric flux lines, and magnetic flux lines are used to describe the magnetic field.
The path of the magnetic flux lines is called the magnetic circuit. Just as a flow of current in the electric circuit
requires the presence of an electromotive force, so the production of magnetic flux requires the presence of
magneto-motive force (mmf ). We now discuss some properties related to magnetic flux.
Flux Density (B) The magnetic flux lines start and end in such a way that they form closed loops.
Weber (Wb) is the unit of magnetic flux (f). Flux density (B) is the flux per unit area. Tesla (T) or Wb/m2 is
the unit of flux density.
f
B = Wb/m 2 or Tesla
A
where B is a quantity called magnetic flux density in teslas, f is the total flux in webers, and A is the area
perpendicular to the lines in m2.
Magneto-motive force, MMF ( ) A measure of the ability of a coil to produce a flux is called the
magneto-motive force. It may be considered as a magnetic pressure, just as emf is considered as an electric
pressure. A coil with N turns which is carrying a current of I amperes constitutes a magnetic circuit and produces an mmf of NI ampere turns. The source of flux (f) in the magnetic circuit is the mmf. The flux produced
in the circuit depends on mmf and the length of the circuit.
Magnetic Field Strength (H) The magnetic field strength of a circuit is given by the mmf per unit
length.
H=
ℑ NI
=
AT/ms
l
l
The magnetic flux density (B) and its intensity (field strength) in a medium can be related by the following
equation
B 5 mH
where
and
m 5 m0 mr is the permeability of the medium in Henrys/metre (H/m),
m0 5 absolute permeability of free space and is equal to 4p 3 10–7 H/m, and
mr 5 relative permeability of the medium.
Relative permeability is a nondimensional numeric which indicates the degree to which the medium is a
better conductor of magnetic flux as compared to free space. The value of mr 5 1 for air and nonmagnetic
materials. It varies from 1,000 to 10,000 for some types of ferromagnetic materials.
Reluctance ( ) It is the property of the medium which opposes the passage of magnetic flux. The
magnetic reluctance is analogous to resistance in the electric circuit. Its unit is AT/Wb. Air has a much higher
reluctance than does iron or steel. For this reason, magnetic circuits used in electrical machines are designed
with very small air gaps.
According to definition, reluctance = mmf
flux
441
Coupled Circuits
1 f
=
ℜ ℑ
Thus, reluctance is a measure of the opposition offered by a magnetic circuit to the setting up of the flux. The
1 l
reluctance of the magnetic circuit is given by ℜ =
.
ma
where l is the length, a is the cross-sectional area of the magnetic circuit and m is the permeability of the
medium.
From the above equations,
1 l ℑ
⋅ =
m a f
The reciprocal of reluctance is known as permeance
ℑ 1 f
= ⋅
1 m a
or
NI
1
= ⋅B
l
m
H=
or
1
⋅B
m
B = mH
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 10
rrr 10-10.1
An iron ring has a mean diameter of 25 cm and a cross-sectional area of 4 cm2. It is wound
with a coil of 1200 turns. An air gap of 1.5 mm width is cut in the ring. Determine the current
required in the coil to produce a flux of 0.48 mwb in the air gap. The relative permeability of
iron under this condition is 800. Neglect leakage.
rrr 10-10.1
Find the net impedance the central mesh and then find the net input impedance of the circuit
shown in Fig. Q.1.
[BPUT 2007]
Frequently Asked Questions linked to LO 10
R1
I1
R1
M12
L1
L2
I2
M23
L3
L 4 I3
R3
Fig. Q.1
10.12
SERIES MAGNETIC CIRCUIT
LO 10
A series magnetic circuit is analogous to a series electric circuit. Kirchhoff’s laws are applicable to magnetic
circuits also. Consider a ring specimen having a magnetic path of l metres, area of cross section (A)m2 with a
mean radius of R metres having a coil of N turns carrying I amperes wound uniformly as shown in Fig. 10.26.
The mmf is responsible for the establishment of flux in the magnetic medium. This mmf acts along the
magnetic lines of force.
442
Circuits and Networks
Fig. 10.26
Flux f 5 mHA webers
The flux produced by the circuit is given by
mmf
f=
reluctance
The magnetic field intensity of the ring is given by
NI
mmf
H=
=
= AT/m
l
l
where l is the mean length of the magnetic path and is given
by 2pR.
NI
Flux density B = m0m r H = m0m r
Wb/m 2
l
NI
× A Wb
l
NI
f=
Wb
l / m 0m r A
= m 0m r
NI is the mmf of the magnetic circuit, which is analogous to emf in electric circuit. l/m0 mr A is the reluctance
of the magnetic circuit which is analogous to resistance in electric circuit.
10.13
COMPARISON OF ELECTRIC AND MAGNETIC CIRCUITS
Series electric and magnetic circuits are shown in Figs. 10.27 (a) and (b) respectively.
LO 11
Figure 10.27 (a) represents an electric circuit with three resistances connected in
series, the dc source E drives the current I through all the three resistances whose
voltage drops are V1, V2 and V3. Hence, E 5 V1 1 V2 1 V3, also E 5 I (R1 1 R2 1 R3).
ρl
We also know that R = , where r is the specific resistance of the material, l is the length of the wire of the
2
resistive material and a is the area of cross section of the wire.
Fig. 10.27
The drop across each resistor V = RI = ρl
or
I
a
V
I
=ρ
l
a
Voltage drop per unit length 5 Specific resistance 3 Current density.
Let us consider the magnetic circuit in Fig. 10.27 (b). The MMF of the circuit is given by ℑ 5 NI, drives
443
Coupled Circuits
1 l
⋅ ,
m a
where l is the length and a is the area of cross section of each arm. The mmf of the magnetic circuit is given
the flux f around the three parts of the circuit which are in series. Each part has a reluctance ℜ =
by ℑ 5 ℑ1 1 ℑ2 1 ℑ3. ℑ 5 f(ℜ1 1 ℜ2 1 ℜ3) where ℜ1, ℜ2 and ℜ3 are the reluctances of the portions 1, 2,
and 3 respectively.
Also
1 l
⋅ ⋅f
m a
ℑ 1 f
= ⋅
m a
l
1
H = ⋅ B.
m
ℑ=
1
can be termed the reluctance of a cubic metre of magnetic material from which, the above equation
m
gives the mmf per unit length (intensity) which is analogous to the voltage per unit length. Parallels between
electric-circuit and magnetic-circuit quantities are shown in Table 10.1.
Thus, it is seen that the magnetic reluctance is analogous to resistance, mmf is analogous to emf, and flux
is analogous to current. These analogies are useful in magnetic-circuit calculations. Though we can draw
many parallels between the two circuits, the following differences do exists.
The electric current is a true flow but there is no flow in a magnetic flux. For a given temperature, r is
independent of the strength of the current, but m is not independent of the flux.
In an electric circuit, energy is expended so long as the current flows, but in a magnetic circuit energy is
expended only in creating the flux, and not in maintaining it. Parallels between the quantities are shown in
Table 10.1.
Table 10.1 Analogy between magnetic and electric circuits
Electric circuit
Magnetic circuit
Exciting force 5 emf in volts
mmf in AT
Response 5 current in amps
f lux in webers
Voltage drop 5 VI volts
V
Electric field density =
volt/m
l
Current ( I ) =
E
A
R
I
Amp/m 2
a
ρl
ohm
a
Conductance (G ) =
Magnetic field Intensity =
ℑ
AT/m
1
ℑ
Web
R
f
Flux density ( B ) = Web/m 2
A
1 l
Reluctance (ℜ ) = ⋅ AT/Web
m a
Flux (f) =
Current density ( J ) =
Resistance ( R) =
mmf drop 5 ℜf AT
1
Mho
R
Permeance =
1 ma l
=
⋅ Web/AT
ℜ
m a
444
Circuits and Networks
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 11
rr 10.11.1
A steel ring of 25 cm mean diameter and of circular section of 3 cm in diameter has an air gap of
1.5 mm length. It is wound uniformly with 700 turns of were carrying a current of 2 A. Calculate
(a) magnetomotive force, (b) flux density, and (c) magnetic flux
Frequently Asked Questions linked to LO 11
r 10-11.1
10.14
Contrast between magnetic circuits and electrical circuits.
[JNTU Nov. 2012]
MAGNETIC LEAKAGE AND FRINGING
Figure 10.28 shows a magnetised iron ring with a narrow air gap, and the flux which
LO 12
crosses the gap can be regarded as useful flux. Some of the total flux produced by
the ring does not cross the air gap, but instead takes a shorter route as shown in
Fig. 10.28 and is known as leakage flux. The flux while crossing the air gap bulges
outwards due to variation in reluctance. This is known
as fringing. This is because the lines of force repel
each other when passing through the air as a result the
flux density in the air gap decreases. For the purpose of
calculation, it is assumed that the iron carries the whole
of the total flux throughout its length. The ratio of total
flux to useful flux is called the leakage coefficient or
leakage factor.
Fig. 10.28
Leakage factor 5 Total flux/Useful flux.
EXAMPLE 10.11
A coil of 100 turns is wound uniformly over a insulator ring with a mean circumference of 2 m and a uniform
sectional area of 0.025 cm2. If the coil is carrying a current of 2 A. Calculate (a) the mmf of the circuit, (b)
magnetic field intensity, (c) flux density, and (d) the total flux.
Solution (a)
mmf 5 NI 5 100 3 2 5 2000 AT
mmf
2000
=
= 1000 AT/m
I
2
(c) B 5 m0H 5 4p 3 10 –7 3 1000 5 1.2565 mWb/m2
(b)
H=
(d) f 5 B 3 A 51.2565 3 10–3 3 0.025 3 10 – 4 5 0.00314 3 10 – 6 Wb
EXAMPLE 10.12
Calculate the mmf required to produce a flux of 5 mWb across an air gap
of 2.5 mm of length having an effective area of 100 cm2 of a cast steel ring
of mean iron path of 0.5 m and cross-sectional area of 150 cm2 as shown
in Fig. 10.29. The relative permeability of the cast steel is 800. Neglect
leakage flux.
Solution Area of the gap 5 100 3 1024 m2
Fig. 10.29
445
Coupled Circuits
−3
4
Flux density of the gap = 5×10 ×10 = 0.5 T
100
B
0.5
H of the gap =
=
m0 4p ×10−7
5 0.39 3 106 A/m
Length of the gap 5 2.5 3 1023 m
mmf required for the gap 5 0.39 3 106 3 2.5 3 10–3 5 975 AT
f
Flux density in the cast steel ring is =
Area
5×10−3 ×10 4
=
100
= 0.333 T
∴ H=
B
0.333
=
= 332 A T/m
m 0m r 4p10−7 × 800
Length of the cast steel path 5 0.5 m
The required mmf for the cast steel to produce the necessary flux 5 0.5 3 332 5 166 AT
Therefore, total mmf 5 975 1 166 5 1141 AT
10.15
COMPOSITE SERIES CIRCUIT
Consider a toroid composed of three different magnetic materials of different
permeabilities, areas and lengths excited by a
coil of N turns.
LO 13
With a current of I amperes as shown
in Fig. 10.30. The lengths of sections AB, BC and CA are l1, l2 and l3
respectively. Each section will have its own reluctance and permeability.
Since all of them are joined in series, the total reluctance of the combined
magnetic circuit is given by
Fig. 10.30
1
mA
l
l
l
= 1 + 2 + 3
m1 A1 m 2 A2 m3 A3
ℜ Total =
The flux produced in the circuit is given by f =






NI
 Wb
∴ f=
 l1
l 
l

+ 2 + 3 
 m1 A1 m 2 A2 m3 A3 
mmf
Wb
Total reluctance
446
10.16
Circuits and Networks
PARALLEL MAGNETIC CIRCUIT
LO 13
We have seen that a series magnetic circuit carries the same flux and the total mmf required to produce a given
quantity of flux is the sum of the mmf’s for the separate parts. In a parallel magnetic circuit, different parts
of the circuit are in parallel. For such circuits the Kirchhoff’s laws, in their analogous magnetic form can be
applied for the analysis. Consider an iron core having three limbs A, B, and C as shown in Fig. 10.31 (a). A
coil with N turns is arranged around limb A which carries a current I amperes. The flux is produced by the
coil in the limb A. fA is divided between limbs B and C and each equal to fA/2. The reluctance offered by the
two parallel paths is equal to the half the reluctance of each path (Assuming equal lengths and cross-sectional
areas). Similar to Kirchhoff’s current law in an electric circuit, the total magnetic flux directed towards a
junction in a magnetic circuit is equal to the sum of the magnetic fluxes directed away from that junction.
Accordingly fA 5 fB 1 fC or fA 2 fB2 fC 5 0. The electrical equivalent of the above circuit is shown in
Fig. 10.31 (b). Similar to Kirchhoff’s second law, in a closed magnetic circuit, the resultant mmf is equal to the
algebraic sum of the products of field strength and the length of each part in the closed path. Thus applying the
law to the first loop in Fig. 10.31 (a), we get
NI 5 HA lA 1 HB lB
or
NI 5 fA ℜA 1 fB ℜB
The mmf across the two parallel paths is identical.
Therefore, NI is also equal to
NI 5 fA ℜA 1 fC ℜC
Fig. 10.31
Frequently Asked Questions linked to LO 13
rrr10-13.1
Two coil with 300 turns and 700 turns are wound side by side on a closed magnetic circuit of area
of 400 cm2 cross section and 80 cm mean length. The magnetic circuit has a relative permeability
of 4000. Determine the mutual inductance, self-induced emf, and mutually induced emf when the
current in the coil with 300 turns grows from zero to 25 A in a time of 0.3 second. [JNTU Nov. 2012]
Additional Solved Problems
PROBLEM 10.1
In the circuit shown in Fig. 10.32, write the equation for the voltages across the coils ab and cd; also mention
the polarities of the terminals.
447
Coupled Circuits
Fig. 10.32
Solution Current i1 is only flowing in coil ab, whereas coil cd is open. Therefore, there is no current in coil
cd. The emf due to self induction is zero on coil cd.
di (t ) with C being positive
∴ v2 (t ) = M 1
dt
Similarly, the emf due to mutual induction in coil ab is zero.
di (t )
∴ v1 (t ) = L 1
dt
PROBLEM 10.2
In the circuit shown in Fig. 10.33, write the equation for the voltages
v1 and v2. L1 and L2 are the coefficients of self-inductances of coils
1 and 2, respectively, and M is the mutual inductance.
Solution In the figure, a and d are like terminals.
Currents i1 and i2 are entering at dot-marked terminals.
di1 (t ) M di2 (t )
+
dt
dt
di (t ) M di1 (t )
v2 = L2 2 +
dt
dt
v1 = L1
Fig. 10.33
PROBLEM 10.3
In Fig. 10.34, L15 4 H; L2 5 9 H, K 5 0.5, i1 5 5 cos (50 t 2 30°) A, i2 5 2
cos (50 t 2 30°) A. Find the values of (a) v1; (b) v2, and (c) the total energy
stored in the system at t 5 0.
Since the current in the coil ab is entering at the dot marked
terminal, whereas in coil cd the current is leaving, we can write the
equations as
Solution
di1
di
−M 2
dt
dt
di
di
v2 = −M 1 + L2 2
dt
dt
v1 = L1
M = K L1 L2 = 0.5 36 = 3
Fig. 10.34
448
(a)
Circuits and Networks
v1 = 4

d 
d
5 cos (50t − 30°) − 3 2 cos (50t − 30°)

dt 
dt
v1 = 20 [− sin (50t − 30°)×50] − 6 [− sin (50t − 30°)50]
at t = 0
(b)
v1 = 500 −150 = 350 V
d
d
v2 = −3 5 cos (50t − 30°) + 9  2 cos (50t − 30°)
dt
dt
v2 = −15 [− sin (50t − 30°)×50] −18 [− sin (50t − 30°)50]
at t = 0
v2 = −375 + 450 = 75 V
(c)
The total energy stored in the system
1
1
2
2
L1 [i1 (t ) ] + L2 [i2 (t ) ] − M [i1 (t )i2 (t ) ]
2
2
1
1
2
2
= × 4 [5 cos (50t − 30°) ] + × 9 [ 2 cos (50t − 30°) ]
2
2
W (t ) =
−3 [5 cos (50t − 30°) × 2 cos (50t − 30°)]
at t = 0 W (t ) = 28.5 j
PROBLEM 10.4
For the circuit shown in Fig. 10.35, write the mesh equations.
Fig. 10.35
Solution There exists mutual coupling between coils 1 and 3, and 2 and 3. Assuming branch currents i1, i2,
and i3 in coils 1, 2, and 3, respectively, the equation for mesh 1 is
v 5 v1 1 v2
v 5 i1 j2 2 i3 j4 1 i2 j4 2 i3 j6
(10.7)
j4i3 is the mutual inductance drop between coils 1 and 3, and is considered negative according to dot
convention and i3 j6 is the mutual inductance drop between coils 2 and 3.
449
Coupled Circuits
For the second mesh, 0 5 2 v2 1 v3
5 2 ( j4i2 2 j6i3) 1 j3i3 2 j6i2 2 j4i1
(10.8)
5 2 j4i1 2 j10i2 1 j9i3
(10.9)
i1 5 i3 1 i2
PROBLEM 10.5
Calculate the effective inductance of the circuit shown in Fig. 10.36
across terminals a and b.
Solution Let the current in the circuit be i.
v=8
di
di
di
di
− 4 + 10 − 4
dt
dt
dt
dt
+5
r
di
di
di
+6 +5
dt
dt
dt
di
di
[34 − 8 ] = 26 = v
dt
dt
Fig. 10.36
di
di
Let L be the effective inductance of the circuit across ab. Then the voltage across ab = v = L = 26
dt
dt
Hence, the equivalent inductance of the circuit is given by 26 H.
PROBLEM 10.6
For the circuit shown in Fig. 10.37, find the ratio of output
voltage to the source voltage.
Solution Let us consider i1 and i2 as mesh currents in the
primary and secondary windings.
As the current i1 is entering at the dot-marked terminal,
and current i2 is leaving the dot-marked terminal, the sign of
the mutual inductance is to be negative. Using Kirchhoff’s
voltage law, the voltage equation for the first mesh is
i1(R1 1 jvL1) 2 i2 jvM 5 v1
i1(10 1 j500) 2 i2 j250 5 10
Fig. 10.37
(10.10)
Similarly, for the second mesh,
i2(R2 1 jvL2) 2 i1 jvM 5 0
i2(400 1 j5000) 2 i1 j250 5 0
(10 + j500) 10
i2 =
− j 250
0
− j 250
(10 + j500)
− j 250
(400 + j5000)
(10.11)
450
Circuits and Networks
i2 5 0.00102 ∠284.13°
v2 5 i2 3 R2
5 0.00102 ∠284.13° 3 400
5 0.408 ∠284.13°
v2 0.408
=
∠− 84.13°
v1
10
v2
= 40.8×10−3 ∠− 84.13°
v1
PROBLEM 10.7
Calculate the effective inductance of the circuit shown in Fig. 10.38
across AB.
Solution The inductance matrix is
L
 11
 L21

L
 31
L12
L22
L32
From KVL,
L13   5
0 −2 
 



L23  =  0
6 −3
L33  −2 −3 17 
v 5 v1 1 v2
Fig. 10.38
(10.12)
and
v2 5 v3
(10.13)
from KCL,
i1 5 i2 1 i3
(10.14)
0 −2  di1 /dt 
 v1   5


  
6 −3  di2 /dt 
 v2  =  0
  
 v3  −2 −3 17   di3 /dt 
and
v1 = 5
di
di1
−2 3
dt
dt
(10.15)
v2 = 6
di
di2
−3 3
dt
dt
(10.16)
v3 = −2
di
di1
di
− 3 2 + 17 3
dt
dt
dt
From Eq. (10.12), we have
v 5 v1 1 v2
(10.17)
Coupled Circuits
=5
v=5
451
di
di
di1
di
− 2 3 + 6 2 −3 3
dt
dt
dt
dt
di
di1
di
+ 6 2 −5 3
dt
dt
dt
(10.18)
From Eq. (10.14),
di1 di2 di3
=
+
dt
dt
dt
(10.19)
Substituting Eq. (10.19) in Eq. (10.17), we have
 di
 di 
 di 
di 
v3 = −2  2 + 3  − 3  2  + 17  3 
 dt
 dt 
 dt 
dt 
or
−5
di
di2
+ 15 3 = v3
dt
dt
(10.20)
Multiplying Eq. (10.16) by 5, we get
30
di
di2
−15 3 = 5v2
dt
dt
Adding Eqs (10.20) and (10.21), we get
di2
= v3 + 5v2
dt
di
25 2 = 6 v2
dt
= 6 v3 , since v2 = v3
25
or v2 =
25 di2
6 dt
From Eq. (10.16),
di
di
25 di2
= 6 2 −3 3
dt
dt
6 dt
from which
di2 18 di3
=
dt
11 dt
From Eq. (10.19),
di1 di2 11 di2 29 di2
=
+
=
dt
dt 18 dt
18 dt
(10.21)
452
Circuits and Networks
di
di2
and 3 in Eq. (10.18) yields
dt
dt
di1
18 di1
11 di2
v = 5 +6
−5
dt
29 dt
18 dt
Substituting the values of
=5
di1 108 di1 55 18 di1
+
−
dt
29 dt 18 29 dt
di
198 di1
= 6.827 1
dt
29 dt
∴ equivalent inductance across AB 5 6.827 H
v=
PROBLEM 10.8
Write the mesh equations for the network shown in Fig. 10.39.
Solution The circuit contains three meshes. Let us assume
three loop currents i1, i2 and i3.
For the first mesh,
5i1 1 j3(i1 2 i2) 1 j4(i3 2 i2) 5 v1
(10.22)
Fig. 10.39
The drop due to self-inductance is j3(i1 2 i2) is written by considering the current (i1 2 i2) entering at
dot-marked terminal in the first coil, j4(i3 2 i2) is the mutually induced voltage in coil 1 due to current (i3 2
i2) entering at dot-marked terminal of the coil 2.
Similarly, for the second mesh,
j3(i2 2 i1) 1 j5(i2 2 i3) 2 j2i2 1 j4(i2 2 i3) 1 j4(i2 2 i1) 5 0
(10.23)
j4(i2 2 i1) is the mutually induced voltage in the coil 2 due to the current in the coil 1, and j4(i2 2 i3) is the
mutually induced voltage in the coil 1 due to the current in the coil 2.
For the third mesh,
3i3 1 j5(i3 2 i2) 1 j4(i1 2 i2) 5 0
(10.24)
Further simplification of Eqs (10.22), (10.23), and (10.24) leads to
(5 1 j3)i1 2 j7i2 1 j4i3 5 v1
2 j7i1 1 j14i2 2 j9i3 5 0
j4i1 2 j9i2 1 (3 1 j5)i3 5 0
(10.25)
(10.26)
(10.27)
PROBLEM 10.9
The inductance matrix for the circuit of three series connected
coupled coils is given in Fig. 10.40. Find the inductances, and
indicate the dots for the coils.
 4 −4 1 


L = −4 2 −3


 1 −3 6 
All elements are in henries.
Fig. 10.40
453
Coupled Circuits
The diagonal elements (4, 2, 6) in the matrix represent
the self-inductances of the three coils 1, 2, and 3, respectively.
The second element in the first row (24) is the mutual inductance
between coils 1 and 2, the negative sign indicates that the current
in the first coil enters the dotted terminal, and the current in
the second coil enters at the undotted terminal. Similarly, the
remaining elements are fixed. The values of inductances and the
dot convention is shown in Fig. 10.41.
Solution
Fig. 10.41
PROBLEM 10.10
Find the voltage across the 10 V resistor for the network
shown in Fig. 10.42.
Solution From Fig. 10.42, it is clear that
v2 5 i2 10
(10.28)
Fig. 10.42
Mesh equation for the first mesh is
j4i1 2 j15 (i1 2 i2) 1 j3i2 5 10 ∠0°
2 j11i1 1 j18i2 5 10 ∠0°
(10.29)
Mesh equation for the second mesh is
j2i2 1 10i2 2 j15(i2 2 i1) 1 j3i1 5 0
j18i1 2 j13i2 1 10i2 5 0
j18i1 1 i2(10 2 j13) 5 0
Solving for i2 from Eqs (10.29) and (10.30), we get
− j11 10∠0°

i2 = 
 j18
0 

∴
− j11
j18 


 j18 10 − j 3


=
−180 ∠90°
291− j110
=
−180 ∠90°
= −0.578 ∠110.7°
311∠20.70°
v2 5 i2 10 5 25.78 ∠110.7°
|v2| 5 5.78
(10.30)
454
Circuits and Networks
PROBLEM 10.11
The resonant frequency of the tuned circuit shown in Fig.
10.43 is 1000 rad/sec. Calculate the self- inductances of the
two coils and the optimum value of the mutual inductance.
Solution From Section 10.7, we know that
1
1
=
L1 C1 L2 C2
1
1
=
=1 H
L1 = 2
2
vr C1 (1000) 1×10−6
1
1
=
= 0.5 H
L2 = 2
2
vr C2 (1000) × 2×10−6
v2r =
Fig. 10.43
Optimum value of the mutual inductance is given by
M optimum =
R1 R2
vr
where R1 and R2 are the resistances of the primary and secondary coils
M=
15
= 3.87 mH
1000
PROBLEM 10.12
The tuned frequency of a double-tuned circuit shown in Fig. 10.44 is 10 4 rad/s. If the source voltage is 2 V
and has a resistance of 0.1 V, calculate the maximum output voltage at resonance if R1 5 0.01 V, L1 52 mH;
R2 5 0.1 V, and L2 5 25 mH.
Fig. 10.44
Solution
The maximum output voltage v0 =
vi
2
2vr C2
Mc
455
Coupled Circuits
where Mc is the critical value of the mutual inductance given by
Mc =
Mc =
vr2 =
At resonance,
C2 =
v0 =
R2 ( R1 + Rs )
vr
0.1(0.01 + 0.1)
10 4
1
L2C2
1
v2r L2
=
= 10.48 mH
1
(104 ) × 25×10−6
2
= 0.4 ×10−3 F
2
( ) ×0.4×10−3 ×10.48×10 −6
2 10
4
2
= 2.385 V
PROBLEM 10.13
An iron ring of 10 cm diameter and 15 cm2 cross section is wound with 250 turns of wire for a flux density
of 1.5 Web/m2 and permeability 500. Find the exciting current, the inductance and stored energy. Find
corresponding quantities when there is a 2 mm air gap.
Solution (a) Without air gap
Length of the flux path 5 pD 5 p 3 10 5 31.41 cm
5 0.3141 m
Area of flux path 5 15 cm2 5 15 3 10 24 m2
mmf 5 A.T
mmf
A=
T
H=
1.5
B
=
= 2387
m 0m r
4p×10−7 ×500
mmf 5 H 3 l 5 2387 3 0.3141 5 750 AT
Exciting current =
Reluctance =
mmf 750
=
= 3A
250
T
1
0.3141
=
−
7
m0m r A 4p10 ×500 ×15×10−4
= 333270
(250)
N2
=
= 0.1875 H
Reluctance 333270
2
Self-inductance =
Energy =
1 2 1
2
LI = × 0.1875×(3)
2
2
= 0.843 j oules
456
Circuits and Networks
(b) With air gap
Reluctance of the gap =
1
2×10−3
=
m0 A 4p×10−7 ×15×10−4
= 1.06×106 A/Wb
Total reluctance 5 (0.333 1 1.06) 106 5 1.393 3 106 A/Wb
mmf 5 f 3 reluctance
5 1.5 3 15 3 1024 3 1.393 3 106
5 3134 AT
Exciting current =
3134
= 12.536 A
250
(250)
N2
=
L=
= 44.8 mH
ℜ
1.393×106
1
Energy = LI 2
2
1
2
= × 44.8×10−3 ×(12.536)
2
= 3.52 joules
2
PROBLEM 10.14
A 700-turn coil is wound on the central limb of the cast steel
frame as shown in Fig. 10.45. A total flux of 1.8 m Wb is required
in the gap. What is the current required? Assume that the gap
density is uniform and that all lines pass straight across the gap.
All dimensions are in centimeters. Assume mr as 600.
Solution Each of the side limbs carry half the total flux as their
Fig. 10.45
reluctances are equal.
Total mmf required is equal to the sum of the mmf required for gap, central limb and side limb.
Reluctance of gap and central limb are in series and they carry the same flux.
Air Gap
fg 5 1.8 3 1023 Wb
Ag 5 4 3 4 3 1024 m2
Bg =
Hg =
1.8×10−3
16 ×10−4
Bg
m0
=
= 1.125 Wb/m 2
1.125
4p×10−7
= 8.95×105 AT/m
Required mmf for the gap 5 Hg lg
5 8.95 3 105 3 0.001 5 895 AT
457
Coupled Circuits
Central Limb
fc 5 1.8 3 1023 Wb
Ac 5 4 3 4 3 1024 m2
Bc 5 1.125 Wb/m2
Hc =
Bc
1.125
=
= 1492 AT/m
m 0m r
4p×10−7 ×600
Required mmf for central limb 5 Hc lc
5 1492 3 0.24 5 358 AT
Side Limb
1
f s = ×flux in central limb
2
1
= ×1.8×10−3 = 0.9×10−3 Wb
2
Bs =
0.9×10−3
= 0.5625 Wb/m 2
16 ×10−4
B
0.5625
Hs = s =
= 746 AT/m
m0m r 4p ×10−7 ×600
Required mmf for side limb 5 Hs ls
5 746 3 0.6 5 447.6 ≅ 448
Total mmf 5 895 1 358 1 448 5 1701 AT
Required current = 1701 = 2.43 A
700
PROBLEM 10.15
Determine i1(t) and i2(t) in the circuit shown in Fig. 10.46. if L1 5 0.4 H;
L2 5 0.4 H; v1 5 15 sin t and M 5 0.2 H.
Solution
di1
di
−M 2
dt
dt
di1
di
15 sint = 0.4 − 0.2 2
dt
dt
di1
di
0 = −M
+ L2 2
dt
dt
di1
di2
0.2
= 0.4
dt
dt
di1
di2
=2
dt
dt
di
di
15 sint = 0.4 × 2 2 − 0.2 2
dt
dt
v1 = L1
Fig. 10.46
458
Circuits and Networks
15ssint = 0.6
di2
dt
di2
= 25 sint
dt
t
i2 (t ) = ∫ 25 sin t dt = −25cost ]t0
0
i2 (t ) = 25(1− cost )
di1
di (t )
= 2 2 ⇒ i1 (t ) = 2i2 (t ) = 50(1− cost )
dt
dt
PROBLEM 10.16
Obtain V1(t) and V2(t) at t 5 2 s for the circuit shown in Fig. 10.47.
If L1 5 0.5 H; L2 5 0.125 H; M 5 0.2 H.
Solution
di1 (t )
di (t )
+M 2
dt
dt
d −2 t
d
= 0.5 (te ) + 0.2 (t 2 e−t )
dt
dt
= 0.5 t (−2)e−2t + e−2t  + 0.2 t 2 (−1)e−t + 2te−t 
V1 (t ) = L1
Fig. 10.47
At t 5 2 s,
V1 (t ) = 0.5  2(−2)e−4 + e−4  + 0.2  22 (−1)e−2 + 2( 2)e−2 
olts
= 0.4056 vo
di2 (t )
di (t )
+M 1
dt
dt
d 2 −t
d
= 0.125 (t e ) + 0.2 (te−2t )
dt
dt
Similarly, V2 (t ) = L2
V2 (t ) = −0.125[ 2te−t − t 2 e−t ] + 0.2 [ e−2t − 2te−2t ]
At t 5 2 s,
( )
( )
V2 (2) = −0.125[ 2 (2) e−2 − 22 e−2 ] + 0.2  e−2 2 − 2 (2) e−2 2 
= −0.01098 volts
PROBLEM 10.17
For the coupled circuit shown in Fig. 10.48, obtain i1(t), i2(t) when the switch is closed at t 5 0 using Laplace
transform.
Coupled Circuits
459
Fig. 10.48
Solution Applying KVL for the first loop when the switch is closed at t 5 0,
1 = Ri1 (t ) + L1
di1 (t )
di (t )
−M 2
dt
dt
Taking Laplace transform of the above equation with zero initial conditions,
1
= I1 (S )[ R + L1S ]− MSI 2 (S )
S
1
= I1 (S )[1 + S ]− 2SI 2 (S )
S
(10.31)
KVL for the second loop
0 = Ri2 (t ) + L2
di2 (t )
di (t )
−M 1
dt
dt
Taking Laplace transform,
= RI 2 (S ) + L2 SI 2 (S ) − MSI1 (S )
= −MSI1 (S ) + ( R + L2 S ) I 2 (S )
Substituting the values,
0 = −2SI1 (S ) + (1 + 4 S ) I 2 (S )
1 + 4S
I1 (S ) = I 2 (S )
2S
Substituting in Eq. (10.31),
(1 + 4S )
1
I 2 (S ) − 2SI 2 (S )
= (1 + S )
S
2S
 (1 + S )

1
(1 + 4S ) − 2S 
= I 2 (S ) 
 2S

S
2
1 + 4 S + S + 4 S − 4 S 2 
1

= I 2 (S ) 

2S
S

(10.32)
460
Circuits and Networks
2
1 + 5S
(1 + 4 S )
and I 1 (S ) = I 2 (S )
2S
2 (1 + 4 S )
=
(1 + 5S ) 2S
1 + 4S
1
1
1
1
I1 (S ) =
= −
= −

(
)
1
S 1 + 5S
S 1 + 5S S
5 S + 

5
Taking inverse Laplace transform of the above equation,
from which
I 2 (S ) =
1
i1 (t ) = t − e−t / 5
5
I 2 (S ) =
=
2
1 + 5S
2/5


S + 1 

5 
t
2 −
i2 (t ) = ⋅ e 5
5
PROBLEM 10.18
An iron ring has a mean circumferential length of 60 cm and a uniform winding of 300 turns. An air gap
has been made by a saw cut across the section of the ring. When a current of 1 A flows through the coil, the
fluxdensity in the air gap is found to be 0.126 m web/m2. How long is the air gap? Assume iron has a relative
permeability of 300. Also calculate the reluctance.
Solution Mean circumferential length 5 60 cm
5 0.6 m
2pr 5 0.6 m
r 5 0.09549 m
Given N 5 300; I 5 1 A; AT 5 300
Cross-sectional area 5 pr2 5 0.02864 m2
Reluctance
R=
0.6
l
=
ma 4p×10−7 ×300 × 0.02864
= 55570.86 AT/wb
Total ampere turns 5 Ampere turns of gap 1 Ampere turns of iron path
461
Coupled Circuits
AT = ATg + ATi
300 =
300 =
Bg l g
m0
+
Bi li
m 0m r
0.126 ×10−3 l g
4p×10−7
from which l g =
where I g -length of air gap
+
0.126 ×10−3 (0.6 − l g )
4p×10−7 ×300
90.345
= 0.3021 metres
299
PROBLEM 10.19
Calculate the current to be passed through the coil ‘C’ having 500 turns so that a flux of 1 m is produced
in the air gap in the Fig. 10.49 shown the core is of square cross section over the entire length and has
permeability of 800.
4 cm
8 cm
I
4 cm
10 cm
24 cm
1 mm
10 cm
4 cm
8 cm
10 cm
10 cm
Fig. 10.49
Solution
Total flux =
.
total mmf
total reluctance
Given figure can be considered as different sections
∴ f=
−3
1×10
=
=
=
mmf
l
l1
l
+ 2 + 3
m1 A1 m 2 A2 m3 A3
I × 500
−3
2
−
(20 + 20)×10
(20 + 8)×10−2
1×10
+
+
−7
−4
−7
−4
−7
−4
4 p ×10 × 64 ×10
800 × 4 p ×10 ×16 ×10
800 × 4 p ×10 × 64 ×10
I ×500 × 4p×10−7
0.3125 + 0.05459 + 2.5
I ×6283×10−7
2.867
I = 4.563 A
462
Circuits and Networks
PSpice Problems
PROBLEM 10.1
Using PSpice, calculate the effective inductance of the
circuit shown in Fig. 10.50 across AB.
* CALCULATION OF EFFECTIVE INDUCTANCE
IS
0
1
AC
10
L1 1
Fig. 10.50
2
L2
2
0
L3 2
3
L4 3
4
R4 01 UOHM
5
6
8
9
K23 L2 L3 20.433
K13 L1 L3 20.316
.AC LIN 1 50 50
.PRINT AC VM(1) VP(1)
.END
**** AC ANALYSIS TEMPERATURE 5 27.000 DEG C
****************************************************************
FREQ
VM(1)
VP(1)
5.000E 1 01 2.145E 1 03
9.000E 1 01
Result
ZEFF 5 XEFF 5 V(1)/IS 5 2145∠90/1∠0 5 j2145
LEFF 5 j2145/(2pf ) 5 6.828 H
Inject 1∠0 current to AB.
After obtaining the driving point impedance across AB, Leff can be calculated.
PSpice will not allow a pure inductor to connect across a voltage source. Hence, inductor voltage source
loop is to be breaked with a small negligible resistance.
Coupled Circuits
PROBLEM 10.2
Using PSpice, for the circuit shown in Fig. 10.51, find the ratio of output voltage to input voltage.
Fig. 10.51
v
2p
= 7.958 Hz
fz =
K12 =
M
L1 L2
=
5
10 ×100
= 0.158 H.
* CALCULATION OF VO / VI
VS 1
0
AC 10 0
R1
1
2
10
L1
2
0
10 H
L2 3
0
100 H
K12 L1 L2 0.158H
R2 3
0
400
.AC LIN 1 7.958 7.958
.PRINT AC VM(R2) VP(R2)
.END
**** AC ANALYSIS
TEMPERATURE 5 27.000 DEG C
****************************************************************
FREQ
VM(R2)
VP(R2)
7.958E 1 00 4.085E 2 01 28.413E 1 01
Result
V2 / V1 5 0.4085 ∠284.13/10 ∠0 5 0.04085 ∠284.13 5 40.85m ∠284.13
463
464
Circuits and Networks
ANSWERS TO PRACTICE PROBLEMS
10-2.3
L 5 13 H
10-2.4
L=
10-2.5
10-3.1
10-5.1
10-6.1
2
M
3
1
Leq = H
3
di
di
di
di
v1 = L1 1 + M 2 ; v2 2 + M 1
dt
dt
dt
dt
10-8.1
1 ∠290 V
10-9.4
I1 5 6.75 2 j0.54; I2 5 23.243 2j0.54
V2
= 0.657 −170.537°
V1
10-9.5

M 2 
Leq =  L1 −
H

L2 
v1 5 181.44 cos (40t 2 30°)
10-10.1 299.6 A
v2 5 202.88 cos (40t 2 30°)
10-11.1 (i) MMR 5 1400 AT
(iii) f 5 0.784 mweb
 2 5 −2 


 5 4 0


−2 0 6 


(ii) B 5 1.120 AT
Objective-Type Questions
rrr 10.1
Mutual inductance is a property associated with
(a) only one coil
(b) two or more coils
(c) two or more coils with magnetic coupling
rrr 10.2
Dot convention in coupled circuits is used
(a) to measure the mutual inductance
(b) to determine the polarity of the mutually induced voltage in coils
(c) to determine the polarity of the self induced voltage in coils
rrr 10.3
Mutually induced voltage is present independently of, and in addition to, the voltage due to self-induction.
(a) true
rrr 10.4
(a) true
rrr 10.5
(b) more than 100%
(c) 90%
The case for which the coefficient of coupling K 5 1 is called perfect coupling is
(a) true
rrr 10.7
(b) false
The maximum value of the coefficient of coupling is
(a) 100%
rrr 10.6
(b) false
Two terminals belonging to different coils are marked identically with dots, if for the different direction
of current relative to like terminals the magnetic flux of self and mutual induction in each circuit add
together.
(b) false
The maximum possible mutual inductance of two inductively coupled coils with self-inductances L1 5 25
mH and L2 5 100 mH is given by
(a) 125 mH
(b) 75 mH
(c) 50 mH
465
Coupled Circuits
rrr 10.8
The value of the coefficient of coupling is more for aircored coupled circuits compared to the iron core
coupled circuits.
(a) true
rrr 10.9
(b) false
Two inductors are connected as shown in Fig. 10.52. What is the value of the effective inductance of the
combination?
(a) 8 H
(b) 10 H
(c) 4 H
Fig. 10.52
rrr 10.10 Two coils connected in series have an equivalent inductance of 3 H when connected in aiding. If the self-
inductance of the first coil is 1 H, what is the self inductance of the second coil (Assume M 5 0.5 H)
(a) 1 H
(b) 2 H
(c) 3 H
rrr 10.11 For Fig. 10.53 shown below, the inductance matrix is given by
3 1


 3 1 2


 1 2 3


(a)  2
−3 1 


−3 1 −2


 1 −2 3 


(b)  2
(c)  2 −3

3

1

Fig. 10.53
For interactive quiz with answers,
scan the QR code given here
OR
visit
http://qrcode.flipick.com/index.php/268
1
2
1

−2
3 
11
LEARNING OBJECTIVES
11.1
STEADY STATE AND TRANSIENT RESPONSE
A circuit having constant sources is said to be in steady state if the currents
and voltages do not change with time. Thus, circuits with currents and voltages
having constant amplitude and constant frequency, sinusoidal functions are also
considered to be in a steady state. That means that the amplitude or frequency of
a sinusoid never changes in a steady state circuit.
LO 1
In a network containing energy storage elements, with change in excitation, the currents and voltages change
from one state to another state. The behaviour of the voltage or current when it is changed from one state to
another is called the transient state. The time taken for the circuit to change from one steady state to another
steady state is called the transient time. The application of KVL and KCL to circuits containing energy storage
elements results in differential, rather than algebraic, equations. When we consider a circuit containing storage
elements which are independent of the sources, the response depends upon the nature of the circuit and is called
the natural response. Storage elements deliver their energy to the resistances. Hence, the response changes with
time, gets saturated after some time, and is referred to as the transient response. When we consider sources
acting on a circuit, the response depends on the nature of the source or sources. This response is called forced
response. In other words, the complete response of a circuit consists of two parts: the forced response and the
transient response. When we consider a differential equation, the complete solution consists of two parts: the
complementary function and the particular solution. The complementary function dies out after a short interval,
and is referred to as the transient response or source free response. The particular solution is the steady state
response, or the forced response. The first step in finding the complete solution of a circuit is to form a differential
equation for the circuit. By obtaining the differential equation, several methods can be used to find out the
complete solution.
467
Transients
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 1*
rrr 11-1.1
What do you understand by transient and steady-state parts of response? How can they be
identified in a general solution?
Frequently Asked Questions linked to LO 1*
rrr 11-1.1
[AU May/June 2013]
[AU May/June 2014]
[RGTU June 2014]
[RGTU June 2014]
Distinguish between natural and forced response.
What is free and forced response?
rrr 11-1.3 What is natural response?
rrr 11-1.4 What do you mean by forced response?
rrr 11-1.2
11.2
DC RESPONSE OF AN R-L CIRCUIT
Consider a circuit consisting of a resistance and inductance as shown in
LO 2
Fig. 11.1. The inductor in the circuit is
initially uncharged and is in series with
the resistor. When the switch S is closed,
we can find the complete solution for the
current. Application of Kirchhoff’s voltage
law to the circuit results in the following differential equation.
Fig. 11.1
V = Ri + L
di
dt
(11.1)
di R
V
(11.2)
+ i=
dt L
L
In the above equation, the current i is the solution to be found and V is the applied constant voltage. The
voltage V is applied to the circuit only when the switch S is closed. The above equation is a linear differential
equation of first order. Comparing it with a non-homogeneous differential equation
dx
+ Px = K
(11.3)
dt
whose solution is
or
(11.4)
x = e− pt ∫ Ke+ Pt dt + ce−Pt
where c is an arbitrary constant. In a similar way, we can write the current equation as
V R /L t
− R /L t
− R /L t
i = ce ( ) + e ( ) ∫ e( ) dt
L
V
−( R /L)t
∴ i = ce
+
(11.5)
R
To determine the value of c in Eq. (11.5), we use the initial conditions. In the circuit shown in Fig. 11.1,
the switch S is closed at t 5 0. At t 5 02, i.e. just before closing the switch S, the current in the inductor is
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
*Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
468
Circuits and Networks
zero. Since the inductor does not allow sudden changes in currents, at t 5 0 just after the switch is closed,
the current remains zero.
Thus, at
t 5 0, i 5 0
Substituting the above condition in Eq. (11.5), we have
0 = c+
c =−
Hence,
V
R
V
R
Substituting the value of c in Eq. (11.5), we get
i=
 R 
V V
− exp − t 
 L 
R R
i=
 R 
V 
1− exp − t 
 L 
R
(11.6)
Equation (11.6) consists of two parts, the steady state part V/R,
and the transient part (V/R)e– (R/L)t. When the switch S is closed, the
response reaches a steady-state value after a time interval as shown
in Fig. 11.2.
Here, the transition period is defined as the time taken for the
current to reach its final or steady state value from its initial value.
In the transient part of the solution, the quantity L/R is important
in describing the curve since L/R is the time required for the
current to reach from its initial value of zero to the final value
 R 
V − L t
is the time at which
e
R
the exponent of e is unity, where e is the base of the natural logarithms. The term L/R is called the time
constant and is denoted by t
L
∴ t=
sec
R
∴ the transient part of the solution is
Fig. 11.2
i =−
V/R. The time constant of a function
 R 
V
V −t /t
exp − t  =
e
 L 
R
R
At one TC, i.e. at one time constant, the transient term reaches 36.8 percent of its initial value.
V
V
V
i( t) = − e−t / t = − e−1 = −0.368
R
R
R
Similarly,
V
V
i( 2t) = − e−2 = −0.135
R
R
V −3
V
i(3 t) = − e = −0.0498
R
R
V
V
i(5 t) = − e−5 = −0.0067
R
R
469
Transients
After 5 TC, the transient part reaches more than 99 percent of its final value.
In Fig. 11.1, we can find out the voltages and powers across each element by using the current.
Voltage across the resistor is
 R 
V
vR = Ri = R × 1− exp − t 
 L 
R 

 R 
∴ vR = V 1− exp − t 

 L 
Similarly, the voltage across the inductance is
vL = L
di
dt
 R 
 R 
V R
= L × exp − t  = V exp − t 
 L 
 L 
R L
Fig. 11.3
The responses are shown in Fig. 11.3.
Power in the resistor is

 R 
 R  V
pR = vR i = V 1− exp − t 1− exp − t 

 L 
 L  R
=
 2 R 
 R 
V 2 
t 
1− 2 exp − t  + exp −
 L 
 L 
R
Power in the inductor is
 R  V
 R 
pL = vL i = V exp − t × 1− exp − t 
 L  R 
 L 
=
Fig. 11.4
 2 R 
 R 
V 2 
t 
exp − t  − exp −
 L 
 L 
R
The responses are shown in Fig. 11.4.
EXAMPLE 11.1
A series RL circuit with R 5 30 V and L 5 15 H has a constant
voltage V 5 60 V applied at t 5 0 as shown in Fig. 11.5. Determine
the current i, the voltage across resistor and the voltage across the
inductor.
Solution By applying Kirchhoff’s voltage law, we get
di
+ 30i = 60
dt
di
+ 2i = 4
∴
dt
15
Fig. 11.5
470
Circuits and Networks
The general solution for a linear differential equation is
i = ce−Pt + e−Pt ∫ Ke Pt dt
where P 5 2, K 5 4
∴ i = ce−2t + e−2t ∫ 4e 2t dt
∴ i 5 ce22t  2
At t 5 0, the switch S is closed.
Since the inductor never allows sudden changes in currents, at t 5 0, the current in the circuit is zero.
Therefore, at t 5 0, i 5 0
∴ 05c2
∴ c522
Substituting the value of c in the current equation, we have
i 5 2(1 2 e22t) A
Voltage across the resistor vR 5 iR
5 2(1 2 e22t) 3 30 5 60 (1 2 e22t) V
di
Voltage across the inductor vL = L
dt
d
= 15× 2(1− e−2t ) = 30 × 2e−2t = 60e−2t V
dt
11.3
DC RESPONSE OF AN R-C CIRCUIT
LO 2
Consider a circuit consisting of resistance and capacitance as shown in Fig. 11.6. The capacitor in the circuit
is initially uncharged, and is in series with a resistor. When the switch
S is closed at t 5 0, we can determine the complete solution for the
current. Application of the Kirchhoff’s voltage law to the circuit results
in the following differential equation.
V = Ri +
Fig. 11.6
1
C
∫ i dt
(11.7)
By differentiating the above equation, we get
0=R
or
di i
+
dt C
1
di
+
i=0
dt RC
(11.8)
(11.9)
Equation (11.9) is a linear differential equation with only the complementary function. The particular
solution for the above equation is zero. The solution for this type of differential equation is
i 5 ce2t/RC
Here, to find the value of c, we use the initial conditions.
(11.10)
471
Transients
In the circuit shown in Fig. 11.6, the switch S is closed at t 5 0. Since the capacitor never allows sudden
changes in voltage, it will act as a short circuit at t 5 0. So, the current in the circuit at t 5 0 is V/R
V
∴ At t = 0, the current i =
R
Substituting this current in Eq. (11.10), we get
V
=c
R
∴ the current equation becomes
V
i = e−t / RC
R
(11.11)
When the switch S is closed, the response decays with time as shown in Fig. 11.7.
In the solution, the quantity RC is the time constant, and is
denoted by t,
where t 5 RC seconds
After 5 TC, the curve reaches 99 percent of its final value. In Fig.
11.6, we can find out the voltage across each element by using the
current equation.
Voltage across the resistor is
V
vR = Ri = R × e−(1/ RC )t ; vR = Ve−t / RC
Fig. 11.7
R
Similarly, voltage across the capacitor is
1
vC = ∫ i dt
C
1 V
= ∫ e−t / RC dt
C R

V
= −
× RC e−t / RC  + c = −Ve−t / RC + c
 RC

At t 5 0, voltage across capacitor is zero
∴ c5V
∴ vC 5 V(1 2 e2t/RC)
The responses are shown in Fig. 11.8.
Power in the resistor
pR = vR i = Ve
−t / RC
V
V 2 −2t / RC
× e−t / RC =
e
R
R
Fig. 11.8
Power in the capacitor
pC = vC i = V (1− e−t / RC )
V −t / RC
e
R
V 2 −t / RC
(e
− e−2t / RC )
R
The responses are shown in Fig. 11.9.
=
Fig. 11.9
472
Circuits and Networks
EXAMPLE 11.2
A series RC circuit consists of a resistor of 10 V and a capacitor
of 0.1 F as shown in Fig. 11.10. A constant voltage of 20 V is
applied to the circuit at t 5 0. Obtain the current equation.
Determine the voltages across the resistor and the capacitor.
Solution By applying Kirchhoff’s law, we get
Fig. 11.10
1
10i +
i dt = 20
0.1 ∫
Differentiating with respect to t, we get
10
i
di
+
=0
dt 0.1
∴
di
+i = 0
dt
The solution for the above equation is i 5 ce2t
At t 5 0, the switch S is closed. Since the capacitor does not allow sudden changes in the voltage, the
current in the circuit is i 5 V/R 5 20/10 5 2 A.
At t 5 0, i 5 2 A.
∴ the current equation i 5 2e2t
Voltage across the resistor is vR 5 i 3 R 5 2e2t 3 10 5 20e2t V
t 

−


Voltage across the capacitor is vC = V 1− e RC 



5 20(1 2 e2t ) V
11.4
DC RESPONSE OF AN R-L-C CIRCUIT
R
s
L
V
+
–
i
C
Fig. 11.11
LO 2
Consider a circuit consisting of resistance, inductance, and capacitance
as shown in Fig. 11.11. The capacitor and inductor are initially
uncharged, and are in series with a resistor. When the switch S is
closed at t 5 0, we can determine the complete solution for the current.
Application of Kirchhoff’s voltage law to the circuit results in the
following differential equation.
V = Ri + L
di 1
+
dt C
∫ idt
(11.12)
By differentiating the above equation, we have
0=R
di
d 2i 1
+L 2 + i
dt
C
dt
(11.13)
Transients
or
d 2i
dt 2
+
R di
1
+
i=0
L dt LC
473
(11.14)
The above equation is a second-order linear differential equation, with only complementary function. The
particular solution for the above equation is zero. Characteristic equation for the above differential equation is

 2 R
 D + D + 1  = 0

L
LC 
(11.15)
The roots of Eq. (11.15) are
D1 , D2 = −
By assuming
K1 = −
R
±
2L
 R 2
  − 1
 2 L 
LC
 R 2
1
R
and K 2 =   −

 2L 
2L
LC
D1 5 K1  K2 and D2 5 K1 2 K2
Here, K2 may be positive, negative or zero.
 R 2
K 2 is positive, when   > 1/LC
 2 L 
The roots are real and unequal, and give the over damped response
as shown in Fig. 11.12. Then Eq. (11.14) becomes
[D – (K1  K2)] [D – (K1 – K2)] i 5 0
The solution for the above equation is
Fig. 11.12
i 5 c1e(K1  K2)t  c2 e(K1 – K2)t
The current curve for the overdamped case is shown in Fig. 11.12.
K2 is negative, when (R/2L)2 , 1/LC
The roots are complex conjugate, and give the underdamped
response as shown in Fig. 11.13. Then Eq. (11.14) becomes
[D – (K1  jK2)] [D – (K1 – jK2)]i 5 0
The solution for the above equation is
i 5 eK1t [c1 cos K2t  c2 sin K2t]
Fig. 11.13
The current curve for the underdamped case is shown in Fig.
11.13.
K2 is zero, when (R/2L)2 5 1/LC
The roots are equal, and give the critically damped response as
shown in Fig. 11.14. Then Eq. (11.14) becomes
(D – K1) (D – K1)i 5 0
The solution for the above equation is
Fig. 11.14
i 5 eK1t(c1  c2t)
The current curve for the critically damped case is shown in Fig. 11.14.
474
Circuits and Networks
EXAMPLE 11.3
The circuit shown in Fig. 11.15 consists of resistance,
inductance, and capacitance in series with a 100 V constant
source when the switch is closed at t 5 0. Find the current
transient.
Fig. 11.15
Solution At the t 5 0, the switch S is closed when the 100 V source is applied to the circuit and results in the
following differential equation.
di
1
100 = 20i + 0.05 +
dt 20 ×10−6
∫ idt
Differentiating Eq. (11.16), we get
0.05
d 2i
dt
2
+ 20
d 2i
dt
∴
2
di
1
+
i=0
dt 20 ×10−6
+ 400
di
+ 106 i = 0
dt
(D2  400D  106)i 5 0
D1 , D2 = −
400
±
2
= −200 ±
 400 2

−106
 2 
2
(200) −106
D1 5 2200  j979.8
D2 5 2200 2 j979.8
Therefore, the current
i 5 ek1t [c1 cos K2t  c2 sin K2t]
i 5 e–200t [c1 cos 979.8t  c2 sin 979.8t] A
At t 5 0, the current flowing through the circuit is zero.
i 5 0 5 (1) [c1 cos 0  c2 sin 0]
∴ c1 5 0
∴ i 5 e–200t c2 sin 979.8t A
Differentiating, we have
di
= c2  e−200t 979.8 cos 979.8t + e−200t (−200) sin 979.8t 


dt
At t 5 0, the voltage across the inductor is 100 V.
(11.16)
Transients
475
di
= 100
dt
di
or
= 2000
dt
di
At t = 0
= 2000 = c2 979.8 cos 0
dt
2000
= 2.04
∴ c2 =
979.8
The current equation is
i 5 e–200t (2.04 sin 979.8t) A
∴
L
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 2
rrr 11-2.1
Obtain an expression for the current i(t) from the differential equation
d 2 i (t )
di(t )
+ 25i(t ) = 0
dt
dt
with initial conditions
di(0+ )
i( 0+ ) = 2
=0
dt
rrr11-2.2 In the given circuit shown in Fig. Q.2, the
switch K is closed at time t 5 0, the steadystate condition having reached previously.
Obtain expression for the current in the circuit
at any time t. If R1 5 R2 5 100 ohms, V 5 10
volts, and L 5 1 henry. Calculate at time
t 5 5 ms, (a) current i, (b) voltage drop across
R2 , and (c) voltage across L.
rrr 11-2.3 In the circuit shown in Fig. Q.3, the capacitor
Fig. Q.2
C has an initial voltage vc(20) 5 10 volts and
at the same instant, the current in the inductor is zero. Switch K is closed at time t 5 0. Obtain
the expression for the voltage v(t) across the inductor L.
2
+ 10
Fig. Q.3
rrr 11-2.4
The network shown in Fig. Q.4 is initially under steady-state condition with the switch in the
position 1. The switch is moved from the position 1 to the position 2 at t  0. Calculate the
476
Circuits and Networks
current i(t) through R1 after switching.
Fig. Q.4
In the network shown in Fig. Q.5, the capacitor c1 is charged to a voltage of 100 V and the
switch S is closed at t 5 0. Determine the current expressions i1 and i2.
rrr 11-2.6 In the circuit shown in Fig. Q.6, the switch K is closed at t 5 0. The current waveform is
observed with CRO. The initial value of the current is measured to be 0.0 1amp. The transient
appears to disappear in 0.1 second. Find the value of R and C.
rrr 11-2.5
Fig. Q.6
Fig. Q.5
rrr 11-2.7
In the circuit shown in Fig. Q.7, steady-state conditions are reached with the switch K in the
position 1. At t 5 0, the switch is changed over to the position 2. Using time domain methods,
determine the current through the inductor i(t) for all t $ 0.
Fig. Q.7
In the circuit shown in Fig. Q.8, the initial
current in the inductance is 2 A and its
direction is as shown in the figure. The
initial charge on the capacitor is 200 C
with polarity as shown when the switch is
closed. Determine the current expression in
the inductance.
rrr 11-2.9 In the circuit shown in Fig. Q.9, the switch
Fig. Q.8
is closed at t 5 0 with zero capacitor
voltage and zero inductor current. Determine V1 and V2 at t 5 0.
rrr 11-2.8
477
Transients
rrr 11-2.10 In
the network shown in Fig. Q.10, determine the current expression for i1(t) and i2(t) when the
switch is closed at t 5 0. The network has no initial energy.
Fig. Q.9
Fig. Q.10
rrr 11-2.11 In
the network shown in Fig. Q.11, the switch is moved from the position 1 to the position 2 at
t 5 0. Determine the current expression.
rrr 11-2.12 Calculate the voltage v1(t) across the inductance for t . 0 in the circuit shown in Fig. Q.12.
Fig. Q.11
Fig. Q.12
Frequently Asked Questions linked to LO 2
rrr 11-2.1
rrr 11-2.2
rrr 11-2.3
rrr 11-2.4
rrr 11-2.5
rrr 11-2.6
rrr 11-2.7
Find the time constant of RL circuit having R = 10 and L = 0.1 mH.
[AU May/June 2013]
A series RL circuit with R = 30 and L = 15H has a
S
constant voltage V = 60 V applied at t = 0 as shown in Fig
30 W
Q.2 Determine the current i, the voltage across resistor
15 H
and the voltage across the inductor.
Iº
60 V
[AU May/June 2014]
What is the time constant of an RL circuit with R = 10
ohms and L = 20 mH?
[AU May/June 2014]
Fig. Q. 2
Derive the transient response of a series R-L circuit with
dc input. Sketch the variation of current and of the voltage across the inductor.
[AU Nov./Dec. 2012]
Solve for i and V as functions of time in the circuit shown in Fig. Q.5, when the switch is closed
at time t = 0.
[AU Nov./Dec. 2012]
Define the term ‘Time constant’ of a circuit, in general.
[AU Nov./Dec. 2012]
In the circuit shown in Fig. Q.7, find the expression for the transient current. The initial current is
as shown in the figure.
[AU April/May 2011]
10 W
i(0)=6A
10 W
3H
+
s
t=0
10 V
i
Fig. Q.5
V 10 mH
i(t)
100 V
Fig. Q.7
5W
478
Circuits and Networks
rrr11-2.8
In the network shown in Fig. Q.8, the switch is closed at t = 0. Find the value of current in each
loop.
[JNTU Nov. 2012]
rrr 11-2.9 In the circuit of Fig. Q.9 the switch is closed at t = 0. Determine the mesh currents i1 (t) and i2 (t).
[AU May/June 2014]
t=0
5W
1H
t=0
1W
100 V
100 W
3H
50 V +
–
i2 (t)
i1 (t)
100 W
2 µF
20 W
Fig. Q.8
Fig. Q.9
rrr 11-2.10
10 W 1
15 W
For the circuit shown in Fig. Q.10 the switch “S” is at
position “I” and the steady-state condition is reached.
t=0
2
The switch is moved to a position “2” at t = 0. Find the
current i(t) in both the cases, i.e., with switch at the 50 V
0.5 H
position I and switch at the position 2.
[GTU Dec. 2010]
rrr 11-2.11 Explain how to determine the initial conditions in
Fig. Q.10
an RL network and the current i(t) based on these
conditions.
[GTU May 2011]
rrr 11-2.12 Using Laplace transformation technique, find current in each loop at t = 0+ following switching at
t = 0 of the switch K as shown in Fig. Q.12. Assume the network previously de-energized.
[JNTU Nov. 2012]
rrr 11-2.13 In the network shown in Fig. Q.13, find the current through the inductor for all values of ‘t’.
[JNTU Nov. 2012]
Fig. Q.12
Fig. Q. 13
rrr 11-2.14
Find i(t) in the circuit of fig. Q.14 when the switch is moved from the position 1 to 2. Initially, it
was under a steady state.
[PU 2010]
rrr 11-2.15 In the circuit shown in Fig. Q.15 the switch is closed at t = 0, here steady state is reached before
t = 0, determine current through the inductor of 3 H.
[PU 2010]
Fig. Q.14
rrr 11-2.16
Fig. Q.15
Fig. Q.16
In the network shown in Fig. Q.16, the voltage source follows the law v(t) = Ve–as, where is a
constant. The switch is closed at t = 0.
[RGTU Dec. 2013]
a) Solve for the current assuming that
=
R
, b) Solve for the current when
L
=
R
L
479
Transients
rrr 11-2.17
Derive the equation for decay of current in R-L circuits. Discuss the role of time constant.
[RGTU June 2014]
rrr 11-2.18 Derive the step responses of RL and RC circuit. Compare their performances.
[AU May/June 2013]
rrr 11-2.19 What is the time constant for RL and RC circuit?
[AU May/June 2014]
rrr 11-2.20 Obtain the response Vc(t) and iL(t) for the source-free RC and RL circuits respectively. Assume
initial voltage Vo and initial current Io respectively.
[GTU Dec. 2010]
rrr 11-2.21 Define the time constant of RL and RC networks and explain the significance of the time
constant.
[GTU May 2011]
rrr 11-2.22 Write down voltage and current relationships in resistor, inductor, and capacitor. Also mention the
initial and final condition for R, L and C components in the different cases.
[GTU Dec. 2012]
rrr11-2.23 In the figure Q.23, the initial voltage in the capacitor is 1 V as the polarity shown. Find the voltage
appearing across with application of the step voltage.
[JNTU Nov. 2012]
rrr 11-2.24 Find UC (t), i1 (t) for t > 0, if the switch is closed at t = 0 after being open for a long time. Refer
Fig. Q.24.
[PU 2010]
t=0
I1
2W
4W
2W
2 u2(t)
3 I2
2W
VC(t)
2W
i1(t)
2F
1F
Fig. Q.23
1F
Fig. Q.24
rrr 11-2.25
What are initial conditions in a network?
[PU 2010]
For the network shown in Fig. Q.26 the switch K is closed at t = 0, with the capacitor uncharged.
Find the values of i, di/dt, d2i/dt2 at t = 0+.
[PU 2012]
rrr 11-2.27 Discuss the initial conditions in a network. Outline the procedure for evaluating the initial
conditions in network problems.
[RGTU Dec. 2013]
rrr 11-2.28 For the circuit shown in the following Fig. Q.28, find the current equation when the switch S is
opened at t = 0+.
[RGTU June 2014]
rrr 11-2.26
10 W
5W
50 V
10 W
2µF
Fig. Q.26
rrr 11-2.29
Fig. Q.28
An RLC series circuit has R = 10 , L = 2 H. What value of capacitance will make the circuit
critically damped?
[AU May/June 2013]
rrr 11-2.31 The circuit shown in Fig. Q.31 consists of resistance, inductance, and capacitance in series with
100 V dc when the switch is closed at t = 0. Find the current transient.
[AU May/June 2013] [BPUT 2008]
rrr 11-2.32 In the circuit shown in Fig. Q.32, the capacitor C has an initial voltage VC = 10 V and at the same
instant when current through the inductor L is zero, the switch K is closed at time t = 0. Find out
the expression for the voltage V(t) across the inductor L.
[BPUT 2008]
480
Circuits and Networks
t=0
R = 5 ohms
t=0
R
20 W
L
0.05
10 V
1F
R = 1/4ohms
L = 1/4 H 3u (t – 3)
V(t)
L=1H
C = 1/4 F
20µF
Fig. Q.31
rrr 11-2.33
Fig. Q.32
Fig. Q.33
Astepvoltage 3u(t–3) is appliedtoa seriesRLCcircuitcomprisinga resistorR= 5 , inductorL= 1H,and
1
capacitor C = F. Find the expression for the current i(t) in the circuit. (Fig. Q.33) [BPUT 2008]
4
rrr 11-2.34 How can volt classify that the given circuit is of first order or second order? Obtain second-order
circuit models for series RLC and parallel RLC circuit in time domain and in “s” domain. [GTU
Dec. 2010]
rrr 11-2.35 In the circuit shown in Fig. Q.35, S1 is closed at t = 0, and S2 is opened at t = 4 msec. Determine
i(t) for t > 0, Assume the inductor is initially de-energized.
[PUT 2011-12]
rrr 11-2.36
In the circuit shown in Fig. Q.36, the switch S is closed at t = 0. Determine the initial value of i,
di/dt, d2ildt2.
[PU 2010]
S1
50 W
S2
t=0
100 V
S
t=0
2H
10 W
t = 4 ms
100 W
i(t)
0.14
1F
100 V
Fig. Q.35
Fig. Q.36
rrr11-2.37
In the circuit shown in Fig. Q.37 switch S is closed at t = 0. Determine the initial value of i, di/dt,
d2ildt2.
[PU 2012]
rrr11-2.38
In the network shown in Fig. Q.38, the switch K is closed at t = 0, with zero capacitor voltage and
zero inductor current.
[RGTU Dec. 2013]
dv1
dv2
and
Solve for (a) v1 and v2 at t = 0, (b) v1 and v2 at t = `
(c)
at t = 0
dt
dt
rrr 11-2.39
Find the current i(t) in a series RLC circuit comprising R = 3 ohms, L = 1 H, and C = 0.5 F, when
a ramp voltage of 10 volts is applied. Assume initial condition as zero.
[RGTU Dec. 2013]
rrr 11-2.40
Figure Q.40 shows a parallel RLC circuit. The switch is suddenly opened at t = 0. Assuming no,
charge on the capacitor and no current in the inductor before switching, find the voltage across the
switch.
[RTU Feb. 2011]
L
R
1 kW
R1
2H
I( )
K
t=0
1
10 V
v
2
C
1mF
Fig. Q.37
L
v1
R2
v2
C
Fig. Q.38
1F
s
Fig. Q.40
1H
1W
481
Transients
11.5
SINUSOIDAL RESPONSE OF AN R-L CIRCUIT
Consider a circuit consisting of resistance and inductance as shown in Fig. 11.16.
LO 3
The switch, S, is closed at t 5 0. At t 5 0, a sinusoidal voltage V cos (vt  u)
is applied to the series R-L circuit,
where V is the amplitude of the
wave and u is the phase angle.
Application of Kirchhoff’s voltage
law to the circuit results in the following differential equation.
Fig. 11.16
V cos (vt + u) = Ri + L
∴
di
dt
(11.17)
di R
V
+ i = cos( vt + u)
dt L
L
The corresponding characteristic equation is


 D + R  i = V cos( vt + u)
(11.18)

L 
L
For the above equation, the solution consists of two parts, viz., complementary function and particular
integral.
The complementary function of the solution i is
ic 5 ce–t(R/L)
The particular solution can be obtained by using undetermined coefficients.
By assuming ip 5 A cos (vt  u)  B sin (vt  u)
i9p 5 – Av sin (vt  u)  Bv cos (vt  u)
Substituting Eqs (11.20) and (11.21) in Eq. (11.18), we have
R
{− Av sin(vt + u) + Bv cos (vt + u)} + { A cos (vt + u)
L
V
+ B sin (vt + u)} = cos (vt + u)
L




V
BR
AR
or − Av +
 sin(vt + u) +  Bv +
 cos (vt + u) = cos (vt + u)


L 
L 
L
Comparing cosine terms and sine terms, we get
BR
− Av +
=0
L
AR V
=
L
L
From the above equations, we have
R
A =V 2
R + ( vL ) 2
Bv +
B =V
vL
R + ( vL ) 2
2
(11.19)
(11.20)
(11.21)
482
Circuits and Networks
Substituting the values of A and B in Eq. (11.20), we get
ip = V
Putting
and
R
R + ( vL )
2
M cos f =
2
cos( vt + u) + V
vL
R + ( vL ) 2
2
sin(vt + u)
(11.22)
VR
R + (v L) 2
vL
M sin f = V 2
R + (v L) 2
2
To find M and f, we divide one equation by the other.
M sin f
vL
= tan f =
M cos f
R
Squaring both equations and adding, we get
M 2 cos 2 f + M 2 sin 2 f =
or
V2
R 2 + ( vL ) 2
M=
V
R + ( vL ) 2
2
∴ the particular current becomes

vL 
V
cos vt + u − tan−1
ip =


2
2
R 
R + ( vL )
(11.23)
The complete solution for the current i 5 ic  ip

vL 
cos vt + u− tan−1


R 
R 2 + ( vL ) 2
Since the inductor does not allow sudden changes in currents, at t 5 0, i 5 0

V
vL 
cos u − tan−1
∴ c =−



R 
R 2 + ( vL ) 2
i = ce−t ( R / L ) +
V
The complete solution for the current is



vL 
−V

cos u − tan−1
i = e−( R / L )t 



R 
 R 2 + ( vL ) 2




vL 
V
+
cos vt + u − tan−1


2
2
R 
R + ( vL )
EXAMPLE 11.4
In the circuit shown in Fig. 11.17, determine the complete solution for the current, when switch S is closed
at t 5 0. Applied voltage is v(t) 5 100 cos (103t  p/2). Resistance R 5 20 V and inductance L 5 0.1 H.
Transients
Fig. 11.17
Solution By applying Kirchhoff’s voltage law to the circuit, we have
20i + 0.1
di
= 100 cos (103 t + p / 2)
dt
di
+ 200i = 1000 cos (1000t + p / 2)
dt
(D  200)i 5 1000 cos (1000t  p/2)
The complementary function ic 5 ce–200t
By assuming particular integral as
ip 5 A cos (vt  u)  B sin (vt  u)
we get

V
vL 
cos vt + u − tan−1
ip =


R 
R 2 + ( vL ) 2
where v 5 1000 rad/s, V 5 100 V
u 5 p/2
L 5 0.1 H, R 5 20 V
Substituting the values in the above equation, we get

100
100 
p
ip =
cos 1000t + − tan−1


2

2
20 
( 20) 2 + 1000 × 0.1
(
=
)


p
100
cos 1000t + − 78.6°



101.9
2


p
= 0.98 cos 1000t + − 78.6°


2
The complete solution is


p
i = ce−200t + 0.98 cos 1000t + − 78.6°


2
At t 5 0, the current flowing through the circuit is zero, i.e., i 5 0
p

∴ c = −0.98 cos  − 78.6°
 2

∴ the complete solution is


p


p
i = −0.98 cos  − 78.6° e−200t + 0.98 cos 1000t + − 78.6°
 2






2
483
484
11.6
Circuits and Networks
SINUSOIDAL RESPONSE OF AN R-C CIRCUIT
LO 3
Consider a circuit consisting of resistance and capacitance in
series as shown in Fig. 11.18. The switch, S, is closed at t 5 0.
At t 5 0, a sinusoidal voltage V cos (vt  u) is applied to the
R-C circuit, where V is the amplitude of the wave and u is the
phase angle. Applying Kirchhoff’s voltage law to the circuit
results in the following differential equation.
V cos (vt + u) = Ri +
R
1
C
Fig. 11.18
∫ idt
(11.24)
di i
+ = −V v sin (vt + u)
dt C


 D + 1  i = − V v sin (vt + u)

RC 
R
(11.25)
The complementary function iC 5 ce–t/RC
(11.26)
The particular solution can be obtained by using undetermined coefficients.
ip 5 A cos (vt  u)  B sin (vt  u)
(11.27)
i9P 5 – Av sin (vt  u)  Bv cos (vt  u)
(11.28)
Substituting Eqs (11.27) and (11.28) in Eq. (11.25), we get
1
{− Av sin (vt + u) + Bv cos (vt + u)} + { A cos (vt + u) + B sin (vt + u)}
RC
Vv
=−
sin (vt + u)
R
B
Vv
Comparing both sides, − Av +
=−
RC
R
A
Bv +
=0
RC
From which,
VR
A=
 1 2
R 2 +  
 vc 
−V

 1 2 
vC  R 2 +   
 vc 


Substituting the values of A and B in Eq. (11.27), we have
and
B=
ip =
VR
2
1
R +  
 vc 
cos (vt + u) +
2
VR
−V

 1 2 
vC  R 2 + 
 vC  


sin(vt + u)
485
Transients

VR
M cos f =
Putting
2

 R 2 +  1  


 vC  

 
 
V
V
M sin f =
and

 1 2 
vC  R 2 + 
 vC  


To find M and f, we divide one equation by the other.
M sin f
1
= tan f =
M cos f
vCR
Squaring both equations and adding, we get
V2
M 2 cos 2 f + M 2 sin 2 f =
2

 R 2 +  1  

 vC  


∴
M=
V
 1 2
R 2 + 

 vC 
The particular current becomes
ip =

1 
cos vt + u + tan−1



v
CR
 1 
2
R + 

 vC 
V
(11.29)
2
The complete solution for the current i 5 ic  ip
∴
− t / RC )
i = ce (
+

1 
cos vt + u + tan−1


vCR 
 1 
2
R + 
 vC 
V
Since the capacitor does not allow sudden changes in voltages at t 5 0, i =
∴
V
cos u = c +
R
c=
(11.30)
2

1 
cos u + tan−1


2

vCR 


1

R 2 + 
 vC 
V
cos u −
R
V

1 
cos u + tan−1


2
vCR 


1

R 2 + 
 vC 
V
V
cosu
R
486
Circuits and Networks
The complete solution for the current is








V
1 
− t / RC ) V
i=e (
cos u + tan−1

 cos u −


2
R
vCR 


1



R 2 + 


 vC 



1 
cos vt + u + tan−1


2
vCR 


1

R 2 + 
 vC 
V
+
(11.31)
EXAMPLE 11.5
In the circuit shown in Fig. 11.19, determine the complete
solution for the current when the switch S is closed


at t 5 0. Applied voltage is v (t ) = 50 cos 102 t + p .

4 
Resistance R 5 10 V and capacitance C 5 1 F.
Fig. 11.19
Solution By applying Kirchhoff’s voltage law to the
circuit, we have
10i +

1
−6
1×10
10
p
∫ idt = 50 cos 100t + 4 

p
di
i
3
+
= −5(10) sin 100t + 
−
6


4
dt 1×10

di
i
p
+ −5 = −500 sin 100t + 


dt 10
4




 D + 1  i = −500 sin 100t + p 

−5 




4
10
The complementary function is iC 5 ce–t/10
(vt  u)  B sin (vt  u),
we get
ip =
. By assuming the particular integral as ip 5 A cos

1 
cos vt + u + tan−1



v
CR
 1 
2
R + 

 vC 
V
where v 5 100 rad/s
C 5 1 F
25
2
u 5 p/4
R 5 10 V
487
Transients
Substituting the values in the above equation, we have
ip =


1
p
cos vt + + tan−1

−6

4
100 ×10 ×10 


1
2
(10) + 

100 ×10−6 
50
2


p
i p = 4.99×10−3 cos 100t + + 89.94°


4
At t 5 0, the current flowing through the circuit is
V
50
cos u =
cos p / 4 = 3.53A
R
10
R
10
V
i = cos u = 3.53 A
R
−5


p
∴ i = ce−t /10 + 4.99×10−3 cos 100t + + 89.94°


4
At t 5 0,
p

c = 3.53 − 4.99×10−3 cos  + 89.94°
 4

Hence, the complete solution is

−5
p

i = 3.53 − 4.99×10−3 cos  + 89.94° e−( t /10 )
 4





p
+ 4.99×10−3 cos 100t + + 89.94°


4
11.7
SINUSOIDAL RESPONSE OF AN R-L-C CIRCUIT
LO 3
Consider a circuit consisting of resistance, inductance, and capacitance in series as shown in Fig. 11.20.
Switch S is closed at t 5 0. At t 5 0, a sinusoidal voltage V
cos (vt  u) is applied to the RLC series circuit, where V is the
amplitude of the wave and u is the phase angle. Application of
Kirchhoff’s voltage law to the circuit results in the following
differential equation.
Fig. 11.20
V cos( vt + u) = Ri + L
di 1
+
dt C
∫ idt
(11.32)
Differentiating the above equation, we get
di
d 2i
R + L 2 + i / C = −V v sin(vt + u)
dt
dt

 2 R
 D + D + 1  i = − V v sin (vt + u)

L
LC 
L
(11.33)
488
Circuits and Networks
The particular solution can be obtained by using undetermined coefficients. By assuming
ip 5 A cos (vt  u)  B sin (vt  u)
i9p 5 – Av sin (vt  u)  Bv cos (vt  u)
ip 5 –
Av2
cos (vt  u) –
Bv2
sin (vt  u)
Substituting ip, i9p and ip in Eq. (11.33), we have
R
− Av2 cos( vt + u) − Bv2 sin(vt + u) + {− Av sin(vt + u) + Bv cos(vt + u)}
L
1
Vv
+
{ A cos(vt + u) + B sin(vt + u)} = − sin(vt + u)
LC
L
(11.34)
(11.35)
(11.36)
}
{
(11.37)
Comparing both sides, we have
Sine coefficients
− Bv 2 − A
vR
B
Vv
+
=−
L
LC
L

 vR 
1  V v
A  + B v2 −
=
 L 

LC 
L
(11.38)
Cosine coefficients
− Av2 + B
A
vR
+
=0
L
LC

 vR 
1 
Av2 −
 − B   = 0

 L 
LC 
Solving Eqs (11.38) and (11.39), we get
V×
A=
B=
v2 R
L2
2
 vR 2 
  − v2 − 1  
 L  
LC  


 2

v − 1 V v

LC 
2
 vR 2 
1  
L   − v2 −



LC  
 L  

Substituting the values of A and B in Eq. (11.34), we get
v2 R
V 2
L
ip =
cos (vt + u)
2
 vR 2 

1
2
  − v −
 
 L  
LC  


(11.39)
Transients
+
 2

v − 1 V v

LC 
2
 vR 2 
1  
L   − v2 −

LC  
 L  

sin(vt + u)
489
(11.40)
v2 R
L2
M cos f =
 vR 2  2
2
  − v − 1 
 L  
LC 
 2
1 
V v −
v

LC 
M sin f =
2
 vR 2 
1  
L   − v2 −

LC  
 L  

V
Putting
and
To find M and f, we divide one equation by the other.
or


vL − 1 


M sin f
vC 
= tan f =
M cos f
R

1  
f = tan−1 vL −
 R

vC  
Squaring both equations and adding, we get
M 2 cos 2 f + M 2 sin 2 f =
∴
M=
V2
 1
2
− vL
R 2 + 
 vC

V
 1
2
− vL
R 2 + 
 vC

The particular current becomes
ip =
V
2
 1

R 2 + 
− vL
 vC


 1



− vL 
 vC



cos  vt + u + tan−1


R




(11.41)
The complementary function is similar to that of a dc series RLC circuit.
To find out the complementary function, we have the characteristic equation

 2 R
 D + D + 1  = 0

L
LC 
(11.42)
490
Circuits and Networks
The roots of Eq. (11.42), are
D1 , D2 =
−R
±
2L
 R 2
  − 1
 2 L 
LC
 R 2
1
R
By assuming K1 = −
and K 2 =   −



2L
2L
LC
∴ D1 5 K1  K2 and D2 5 K1 2 K2
K2 becomes positive, when (R/2L)2 . 1/LC
The roots are real and unequal, which gives an overdamped response. Then Eq. (11.42) becomes
[D 2 (K1  K2)] [D 2 (K1 2 K2)]i 5 0
The complementary function for the above equation is
ic = c1e( K1 + K 2 )t + c2 e( K1 −K 2 )t
Therefore, the complete solution is
i 5 ic  ip
ic = c1e
( K1 + K 2 )t
+ c2 e( K1 −K 2 )t +

 1
vL 

cos  vt + u + tan−1 
−
 vCR R 

2



1
R 2 + 
− vL
 vC

V
 2
K2 becomes negative, when  R  < 1
 2 L 
LC
Then the roots are complex conjugate, which gives an underdamped response. Equation (11.42) becomes
[D 2 (K1  jK2)] [D 2 (K1 2 jK2)]i 5 0
The solution for the above equation is
ic 5 eK1t [c1 cos K2t  c2 sin K2t]
Therefore, the complete solution is
i 5 ic  ip
i 5 eK1t [c1 cos K2t  c2 sin K2t]
+

 1
vL 
cos  vt + u + tan−1 
− 
 vCR R 



 1

R 2 + 
− vL
 vC

V
2
 R 2
K2 becomes zero, when   = 1 / LC
 2 L 
Transients
491
Then the roots are equal which gives critically damped response. Then, Eq. (11.42) becomes (D – K1)
(D – K1)i 5 0.
The complementary function for the above equation is
ic 5 eK1t (c1  c2t)
Therefore, the complete solution is i 5 ic  ip
∴ i 5 eK1t[c1  c2t] +

 1
vL 
cos  vt + u + tan−1 
− 
 vCR R 


 1
2
R 2 + 
− vL
 vC

V
EXAMPLE 11.6
In the circuit shown in Fig. 11.21, determine the complete solution for the current, when the switch is closed
p
at t 5 0. Applied voltage is v (t ) = 400 cos 500t + . Resistance R 5 15 V, inductance L 5 0.2 H, and


4
capacitance C 5 3 F.
Fig. 11.21
Solution By applying Kirchhoff’s voltage law to the circuit,
15i(t ) + 0.2
di(t )
1
+
dt
3×10−6
∫

p
i(t )dt = 400 cos 500t + 

4
Differentiating the above equation once, we get
15

p
di
d 2i
i
+ 0.2
+
= −2×105 sin 500t + 

4
dt
dt 3×10−6

p
−2×105
sin 500t + 

0.2
4
The roots of the characteristic equation are
D1 5 237.5  j1290 and D2 5 237.5 2 j1290
The complementary current
ic 5 e–37.5t (c1 cos 1290t  c2 sin 1290t)
Particular solution is

 1
vL 
V
cos  vt + u + tan−1 
ip =
− 


 vCR R 
 1
2
R 2 + 
− vL
 vC

( D 2 + 75 D + 16.7×105 )i =


p
∴ i p = 0.71 cos 500t + + 88.5°


4
492
Circuits and Networks
The complete solution is
i 5 e–37.5t (c1 cos 1290t  c2 sin 1290t)  0.71 cos (500t  45°  88.5°)
At t 5 0, i0 5 0
∴ c1 5 20.71 cos (133.5°) 5  0.49
Differentiating the current equation, we have
di
= e−37.5t (−1290c1 sin 1290t + c2 1290 cos 1290t )
dt
237.5e237.5t(c1 cos 1290t  c2 sin 1290t)
20.71 3 500 sin (500t  45°  88.5°)
di
= 1414
dt
∴ 1414 5 1290c2 2 37.5 3 0.49 2 0.71 3 500 sin (133.5°)
1414 5 1290c2 2 18.38 2 257.5
∴ c2 5 1.31
The complete solution is
i 5 e–37.5t (0.49 cos 1290t  1.31 sin 1290t)  0.71 cos (500t  133.5°)
At t = 0,
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 3
In the network shown in Fig. Q.1, find i2(t) for t . 0, if i1(0) 5 5 A.
rrr 11-3.2 The switch in Fig. Q.2 was open for a long time but closed at t 5 0. If i (0) 5 10 A, find i (t) for
t . 0 by hand and also PSpice.
rrr 11-3.1
Fig. Q.1
rrr 11-3.3
rrr 11-3.4
Fig. Q.2
Using PSpice, find V (t) for t , 0 and t . 0 in the circuit shown in Fig. Q.3.
In the network shown in Fig. Q.4, the switch is closed at t 5 0 and there is no initial charge on
either of the capacitances. Find the resulting current i(t).
Fig. Q.3
Fig. Q.4
Transients
rrr 11-3.5
493
In the RC circuit shown in Fig. Q.5, the capacitor has an initial charge q0 5 25 3 1026 C with
polarity as shown. A sinusoidal voltage v 5 100 sin (200t  f) is applied to the circuit at a time
corresponding to f 5 30°. Determine the expression for the current i(t).
Fig. Q.5
rrr11-3.6
For the circuit shown in Fig. Q.6, find v5, if the switch is opened for t . 0.
Fig. Q.6
rrr 11-3.7
Determine the response V (t) using PSpice for the circuit shown in Fig. Q.7.
Fig. Q.7
rrr 11-3.8
1
1
In the network shown in Fig. Q.8, the values of R, L, and C are ohm, H and 1 F respectively.
4
4
If I 5 5 amp and the switch K is opened at t 5 0. Obtain an expression for voltage of the node
a for t $ 0.
Fig. Q.8
494
Circuits and Networks
Frequently Asked Questions linked to LO 3
r 11-3.1
rr 11-3.2
rr 11-3.3
rr 11-3.4
An RL series circuit is excited by a sinusoidal source e(t) = 10 sin 100t volts, by closing the switch
at t = 0. Take R = 10 and L = 0.1 H. Determine the current i(t) flowing through the RL circuit.
[AU May/June 2014]
100 W
2 K
Obtain the current at t > 0, if ac voltage V is
applied when the switch K is moved to 2 from
1
1 at t = 0. Assume a steady-state current of 1 V = 100
0.1 H
A in LR circuit when the switch was at the sin 314 t
position 1.
[RTU Feb. 2011]
Derive the expression for the complete
Fig. Q.2
solution of the current response of an RC
series circuit with an excitation of V cos( t + f). Briefly explain the significance of phase angle
in the solution.
[AU Nov./Dec. 2012]
Derive the expression for the complete solution of the current response of an RC series circuit with
an excitation of V cos( t + f). Briefly explain the significance of phase angle in the solution.
[AU April/May 2011]
V
Write down the condition for critically damped response of a series
RLC circuit excited by a sinusoidal ac source.
[AU Nov./Dec. 2012]
I
I
R
In the network shown in Fig. Q.6, the switch k is closed at t = 0,
R
+
–
connecting voltage Vo sin t to the parallel RL- RC circuit. Find
Vasinwt
L
C
(a) di1/dt and (b) di2/ dt at t = 0.
[JNTU Nov. 2012]
Plot the response of RLC circuit to sinusoidal input. [RGTU June 2014]
Derive an expression for the current response of an RLC series circuit
Fig. Q.6
with sinusoidal excitation. Assume that the circuit is working in critical
damping condition.
[AU May/June 2013]
1
r 11-3.5
rrr11-3.6
rr 11-3.7
rr 11-3.8
1
2
Additional Solved Problems
PROBLEM 11.1
A series circuit shown in Fig. 11.22 comprising of a resistance of
10 V and an inductance of 0.5 H, is connected to a 100 V source
at t 5 0. Determine the complete expression for the current i(t).
Solution The current equation for the circuit is shown in Fig.
11.22 by using Kirchhoff’s voltage law when the switch is
closed.
di(t )
10i(t ) + 0.5
= 100
dt
di(t )
+ 20 i(t ) = 200
dt
The general solution for a linear differential equation is
i(t ) = Ce− pt + e− pt ∫ Ke pt dt
where p 5 20, K 5 200
Fig. 11.22
495
Transients
∴ i(t ) = Ce−20t + e−20t ∫ 200 e 20t dt
i(t ) = Ce−20t + 10
At t 5 0, the switch S is closed.
Since the inductor never allows sudden changes in currents, At t 5 0, the current in the circuit is zero.
∴
At t 5 0, i 5 0
∴
0 5 C  10
∴
C 5 210
Substituting the value of C in the current equation, we have
i 5 10(1 2 e220t) A
PROBLEM 11.2
A series RLC circuit shown in Fig 11.23, comprising R 5 10 V,
L 5 0.5 H, and C 5 1 F, is excited by a constant voltage source
of 100 v. Obtain the expression for the current. Assume that the
circuit is relaxed initially.
Solution At t 5 0, the switch S is closed when the 100 V source
is applied to the circuit and results in the following differential
equation:
100 = 10i + 0.5
di
1
+
dt 1×10−6
∫ idt
Differentiating the above equation, we get
0.5
d 2i
dt
2
+ 10
d 2i
dt
2
di
1
+
i=0
dt 1×10−6
+ 20
di
+ 2×106 i = 0
dt
( D 2 + 20 D + 2×106 )i = 0
D1 , D2 =
−20
±
2
 20 2
  − 2×106
 2 
2
= −10 ± (10) − 2×106
D1 = −10 + j1414.2
D2 = −10 − j1414.2
Therefore, the current
= e k1t [ c1 cos K 2 t + c2 sin K 2 t ]
i = e−10t [ c1 cos 1414.2t + c2 sin 1414.2t ] A
At t 5 0, the current flowing through the circuit is zero.
Fig. 11.23
496
Circuits and Networks
i = 0 = (1) [ c1 cos 0 + c2 sin 0 ]
c1 = 0
∴ i = e−10t c2 sin 1414.2t A
Differentiating, we have
di
= c2  e−10t 1414.2 cos 1414.2t + e−10t (−10) sin 1414.2t 
dt
At t 5 0, the voltage across the inductor is 100 V.
L
di
= 100
dt
di 100
=
= 200
dt
0.5
di
At t = 0
= 200 = c2 ×1414.2 cos 0
dt
200
= 0.1414
c2 =
1414.2
or
The current equation is
i = e−10t (0.1414 sin 1414.2t )A
PROBLEM 11.3
In the network shown in Fig. 11.24, the switch is moved from the position 1 to the position 2 at t 5 0. The
switch is in position 1 for a long time. Determine the current expression i(t).
Fig. 11.24
Solution At t 5 0, the switch S is moved to the position 2 and the 50 V source is applied to the circuit and
results in the following differential Eq. (11.43).
50 = 10i + 2
di 1
q(0)
+ ∫ idt +
dt 3
3
(11.43)
Transients
497
Differentiating Eq. (11.43), we get
2
d 2i
di 1
+ 10 + i = 0
dt
dt 3
d 2i
di 1
+5 + i = 0
dt
dt 6
 2

 D + 5 D + 1  i = 0

6 
D1 , D2 =
−5
±
2
 5 2  1 
  −  
 2   6 
= −2.5 ± 6.25 − 0.167
=− ±
= −2.5 ± 2.47
[ D − (−0.03)][ D − (−4.97)]i = 0
The solution for the above equation is
i = c1e−0.03t + c2 e−4.97t
At t 5 0 , i 5 0
∴ c1  c2 5 0
di
= −0.03c1e−0.03t − 4.97c2 e−4.97t
dt
At t 5 0, the voltage across the inductor is 50 V.
L
(11.44)
di
= 50
dt
di
= 25
dt
At t 5 0, Eq. (11.44) becomes
25 = −0.03 c1 − 4.97 c2
∴ c1 = 5.06, c2 = −5.06
The current equation is
i = 5.06 e−0.03t − 5.06 e−4.97t
PROBLEM 11.4
In the network in Fig. 11.25, the switch is moved from the position 1 to the position 2 at t 5 0. The switch is
in the position 1 for a long time. Initial charge on the capacitor is 7 3 1024 coulombs. Determine the current
expression i(t), when v 5 1000 rad/s.
498
Circuits and Networks
Fig. 11.25
Solution When the switch is at the position 2, by applying Kirchhoff’s law, the differential equation is
100i +
100i +
1
20 ×10−6
1
20 ×10−6
t
∫ i dt = 0
0
∫ i dt +v(0) = 0
Where v(0) = initial voltage across capacitor
1
7×10−4
100i +
i
dt
+
=
∫
20 ×10−6
20 ×10−6
(11.45)
Differentiating Eq. (11.45), we get
di
1
+
i=0
dt 20 ×10−6


1
 D +
i = 0
−
4

20 ×10 
100
The transient current is
−1
t
−4
20
×
= c e 10
i
i = c e−50t
At t 5 0, the switch is moved from the position 1 to the position 2. Hence, the current passing through the
circuit is same as the steady-state current passing through the circuit, when the switch is in the position 1.
=
i=
100 30°
100 30°
v
=
=
z R − j / vc 50 − j 50
100 30°
70.71 −45°
i = 1.414 75° A
Therefore, the steady-state current passing through the circuit, when the switch is in the position 1 is
= 1.414 sin (1000t + 75°) A
t t = 0 ; i = 1.414 sin 75° = 1.365 A
Therefore, the current equation is i 5 1.365 e250t A
Transients
499
PROBLEM 11.5
The switch in the circuit shown in Fig. 11.26 is closed at t 5 0. Find v2(t) for all t $ 0 by time domain method.
Assume zero initial current in the inductance.
Fig. 11.26
Solution By applying KVL to the loop 1, we get
10 = 30 [i1 (t ) + i2 (t ) ] + 10 i1 (t )
40 i (t ) + 30 i (t ) = 10
By applying KVL to the loop 2, we get
0 i1 (t ) + i2 (t ) + 0.2
di2 (t )
= 10
dt
(11.46)
(11.47)
From Eq. (11.46),
i(t) 5 0.25 – 0.75i2(t)
(11.48)
Substituting Eq. (11.48) into Eq. (11.47), we get
di2 (t )
− 37.5 i2 (t ) = 2.5
dt
( D − 37.5) i2 = 2.5
(11.49)
i2(t) 5 0.066  c e237.5t
At t 5 0 ; i2(t) 5 0
c 5 20.066
∴
i2(t) 5 0.066[1 – e237.5t] A
di2 (t )
dt
d
= 0.2  0.066(1− e−37.5t )
dt
v2 ( t ) = L
v2 (t ) = 0.495 e−37.5t V
PROBLEM 11.6
The circuit shown in Fig. 11.27, consists of series RL elements with R 5 150 V and L 5 0.5 H. The switch is
closed when f 5 30°. Determine the resultant current when voltage V 5 50 cos (100t  f) is applied to the
circuit at f 5 30°.
500
Circuits and Networks
Fig. 11.27
Solution By using Kirchhoff’s laws, the differential equation, when the switch is closed at f 5 30°, is
150i + 0.5
di
= 50 cos (100t + f)
dt
0.5Di  150i 5 50 cos (100t  30°)
(D  300)i 5 100 cos (100t  30°)
The complementary current ic 5 ce–300t
To determine the particular current, first we assume a particular current.
ip 5 A cos (100t  30°)  B sin (100t  30°)
Then
i9p 5 – 100 A sin (100t  30°)  100 B cos (100t  30°)
Substituting ip and i9p in the differential equation and equating the coefficients, we get
– 100 A sin (100t  30°)  100 B cos (100t  30°)  300 A cos (100t  30°)
 300B sin (100t  30°) 5 100 cos (100t  30°)
–100 A  300 B 5 0
300 A  100 B 5 100
From the above equation, we get
A 5 0.3 and B 5 0.1
The particular current is
ip 5 0.3 cos (100t  30°)  0.1 sin (100t  30°)
∴ ip 5 0.316 cos (100t  11.57°) A
The complete equation for the current is i 5 ip  ic
∴ i 5 ce–300t  0.316 cos (100t  11.57°)
At t 5 0, the current i0 5 0
∴ c 5 – 0.316 cos (11.57°) 5 – 0.309
Therefore, the complete solution for the current is
i 5 – 0.309e–300t  0.316 cos (100t  11.57°) A
PROBLEM 11.7
The circuit shown in Fig. 11.28, consists of series RC
elements with R 5 15 V and C 5 100 F. A sinusoidal
voltage v 5 100 sin (500t  f) volts is applied to the
circuit at time corresponding to f 5 45°. Obtain the
current transient.
Fig. 11.28
Transients
Solution By using Kirchhoff’s laws, the differential equation is
1
∫ idt = 100 sin (500t + f)
100 ×10−6
Differentiating once, we have
15i +
15
di
1
i = (100) (500) cos (500t + f)
+
dt 100 ×10−6

1
 i = 3333.3 cos (500t + f)
 D +
−6 

1500 ×10 
(D  666.67)i 5 3333.3 cos (500t  f)
The complementary function ic 5 ce–666.67t
To determine the particular current, first we assume a particular current.
ip 5 A cos (500t  45°)  B sin (500t  45°)
i9p 5 – 500 A sin (500t  45°)  500 B cos (500t  45°)
Substituting ip and i9p in the differential equation, we get
– 500 A sin (500t  45°)  500 B cos (500t  45°)
 666.67 A cos (500t  45°)  666.67 B sin (500t  45°)
5 3333.3 cos (500t  f)
By equating coefficients, we get
500 B  666.67 A 5 3333.3
666.67 B – 500 A 5 0
From which, the coefficients
A 5 3.2; B 5 2.4
Therefore, the particular current is
ip 5 3.2 cos (500t  45°)  2.4 sin (500t  45°)
ip 5 4 sin (500t  98.13°)
The complete equation for the current is
i 5 ic  ip
i 5 ce–666.67t  4 sin (500t  98.13°)
At t 5 0, the differential equation becomes
15i 5 100 sin 45°
100
i=
sin 45° = 4.71 A
15
∴ at t 5 0,
4.71 5 c  4 sin (98.13°)
∴ c 5 0.75
The complete current is
i 5 0.75 e–666.67t  4 sin (500t  98.13°)
501
502
Circuits and Networks
PROBLEM 11.8
The circuit shown in Fig. 11.29 consisting of series RLC
elements with R 5 10 V, L 5 0.5 H and C 5 200 F
has a sinusoidal voltage v 5 150 sin (200t  f). If the
switch is closed when f 5 30°, determine the current
equation.
Fig. 11.29
Solution By using Kirchhoff’s laws, the differential
equation is
10i + 0.5
di
1
+
dt 200 ×10−6
∫ idt = 150 sin (200t + f)
Differentiating once, we have
(D2  20D  104)i 5 60000 cos (200t  f)
The roots of the characteristics equation are
D1 5 210  j99.49 and D2 5 210 2 j99.49
The complementary function is
ic 5 e–10t (c1 cos 99.49t  c2 sin 99.49)
We can find the particular current by using the undetermined coefficient method.
Let us assume
ip 5 A cos (200t  30°)  B sin (200t  30°)
i9p 5 – 200 A sin (200t  30°)  200 B cos (200t  30°)
ip 5 – (200)2 A cos (200t  30°) – (200)2 B sin (200t  30°)
Substituting these values in the equation, and equating the coefficients, we get
A 5 0.1 B 5 0.067
Therefore, the particular current is
ip 5 1.98 cos (200t – 52.4°) A
The complete current is
i 5 e–10t (c1 cos 99.49t  c2 sin 99.49t)  1.98 cos (200t – 52.4°) A
di
= 300
From the differential equation at t = 0, i0 = 0 and
dt
∴ At t 5 0
c1 5 21.98 cos (252.4°) 5 21.21
Differentiating the current equation, we have
di
= e−10t (−99.49c1 sin 99.49t + 99.49c2 cos 99.49t )
dt
– 200 (1.98) sin (200t – 52.4°) – 10e–10t (c1 cos 99.49t  c2 sin 99.49t)
503
Transients
At t = 0,
di
= 300 and c1 = −1.21
dt
300 5 99.49 c2 2 396 sin (252.4°) 2 10(21.21)
300 5 99.49 c2  313.7  12.1
c2 5 225.8
Therefore, the complete current equation is
i 5 e –10t (0.07 cos 99.49t 2 25.8 sin 99.49t)  1.98 cos (200t 2 52.4°) A
PROBLEM 11.9
For the circuit shown in Fig. 11.30, determine the
transient current when the switch is moved from
the position 1 to the position 2 at t 5 0. The circuit
is in steady state with the switch in position 1.
The voltage applied to the circuit is v 5 150 cos
(200t  30°) V.
Fig. 11.30
Solution When the switch is at the position 2, by applying Kirchhoff’s law, the differential equation is
200i + 0.5
di
=0
dt
(D  400)i 5 0
∴ the transient current is
i 5 ce – 400t
At t 5 0, the switch is moved from the position 1 to the position 2. Hence, the current passing through
the circuit is the same as the steady state-current passing through the circuit when the switch is in position 1.
When the switch is in the position 1, the current passing through the circuit is
=
v 150 ∠30°
=
z R + j vL
=
150 ∠30°
150 ∠30°
=
= 0.67 ∠3.44°
200 + j (200) (0.5) 223.6 ∠26.56°
Therefore, the steady-state current passing through the circuit when the switch is in the position 1 is
i 5 0.67 cos (200t  3.44°)
Now substituting this equation in the transient current equation, we get
0.67 cos (200t  3.44°) 5 ce – 400t
At t 5 0; c 5 0.67 cos (3.44°) 5 0.66
Therefore, the current equation is i 5 0.66e – 400t
504
Circuits and Networks
PROBLEM 11.10
In the circuit shown in Fig. 11.31, determine the current
equations for i1 and i2 when the switch is closed at t 5 0.
Solution By applying Kirchhoff’s laws, we get two equations:
35i1  20i2 5 100
di
20i1 + 20i2 + 0.5 2 = 100
dt
From Eq. (11.50), we have
(11.50)
Fig. 11.31
(11.51)
35i1 5 100 – 20i2
100 20
i1 =
− i2
35 35
Substituting i1 in Eq. (11.51), we get
100 20 
di
20 
− i + 20i2 + 0.5 2 = 100
 35 35 2 
dt
57.14 −11.43i2 + 20i2 + 0.5
(11.52)
di2
= 100
dt
(D  17.14)i2 5 85.72
From the above equation,
i2 5 ce –17.14t  5
Loop current i2 passes through inductor and must be zero at t 5 0
At t 5 0, i2 5 0
∴ c5–5
∴ i2 5 5(1 – e –17.14t) A
and the current i1 5 2.86 – {0.57 3 5(1 – e –17.14t)}
5 (0.01  2.85 e –17.14t) A
PROBLEM 11.11
For the circuit shown in Fig. 11.32, find the current equation when
the switch is changed from the position 1 to the position 2 at t 5 0.
Solution By using Kirchhoff’s voltage law, the current equation
is given by
Fig. 11.32
0i + 0.4
di
= 10i
dt
505
Transients
At t 5 0 –, the switch is at the position 1, the current passing through the circuit is
500
i(0− ) =
= 5A
100
di
0.4 + 50i = 0
dt


 D + 50  i = 0

0.4 
i = ce−125t
At t 5 0, the initial current passing through the circuit is same as the current passing through the circuit
when the switch is at the position 1.
At t 5 0, i(0) 5 i(0 –) 5 5 A
At t 5 0, c 5 5 A
∴ the current I 5 5e –125t
PROBLEM 11.12
For the circuit shown in Fig. 11.33, find the current equation
when the switch S is opened at t 5 0.
Solution When the switch is closed for a long time,
At t = 0− , the current i(0− ) =
100
= 5A
20
Fig. 11.33
When the switch is opened at t 5 0, the current equation by using Kirchhoff’s voltage law is given by
1
∫ i dt +10i = 5i
4 ×10−6
1
∫ i dt + 5i = 0
4 ×10−6
Differentiating the above equation,
di
1
+
i=0
dt 4 ×10−6

1
D+
i = 0
−6 
20 ×10 
5
∴
i
−1
t
−6
20
×
= ce 10
At t 5 0–, just before the switch S is opened, the current passing through the 10 V resistor is 5 A. The same
current passes through 10 V at t 5 0.
∴ At t 5 0, i(0) 5 5 A
At t 5 0, c1 5 5 A
−t
−6
The current equation is = 5e 20×10
506
Circuits and Networks
PROBLEM 11.13
For the circuit shown in Fig. 11.34, find the current in the 20 V
when the switch is opened at t 5 0.
Solution When the switch is closed, the loop currents i1 and i2 are
flowing in the circuit.
The loop equations are 30(i1 – i2)  10i2 5 50
30(i2 – i1)  20i2 5 10i2
Fig. 11.34
From the above equations, the current in the 20 V resistor i2 5 2.5 A.
The same initial current is flowing when the switch is opened at t 5 0.
When the switch is opened, the current equations
di
= 10i
dt
2di
=0
40i +
dt
0i + 20i + 2
(D  20)i 5 0
i 5 ce–20t
At t 5 0, the current i(0) 5 2.5 A
∴ At t 5 0, c 5 2.5
The current in the 20 V resistor is i 5 2.5 e –20t
PROBLEM 11.14
For the circuit shown in Fig. 11.35, find the current equation when the switch is opened at t 5 0.
Fig. 11.35
Solution When the switch is closed, the current in the 20 V resistor i can be obtained using Kirchhoff’s
voltage law.
10i  20i  20i 5 100
50i 5 100, ∴ i 5 2 A
The same initial current passes through the 20 V resistor when the switch is opened at t 5 0.
The current equation is
20i + 10i +
10i +
1
2×10−6
1
2×10−6
∫ idt = 20i
∫ idt = 0
Transients
507
Differentiating the above equation, we get
di
1
+
i=0
dt 2×10−6


1
 D +
i = 0
−6 

20 ×10 
10
The solution for the above equation is
−1
−6
i = ce 20×10
t
At t 5 0, i(0) 5 i(0 –) 5 2 A
∴ At t 5 0, c 5 2 A
The current equation is
−1
−6
i = 2e 20×10
t
PSpice Problems
PROBLEM 11.1
Using PSpice, for the circuit shown in Fig. 11.36, find the complete expression for the current when the switch
is closed at t 5 0.
Fig. 11.36
RL TRANSIENT WITH SWITCH
VS
1
0
PWL(0 100 1N 100 1U 100 1 100)
R1
1
2
20
R2
2
3
30
L3 0 0.1H IC 5 2
W1 2 3 VS SMOD
.MODEL SMOD ISWITCH(ION 5 0 IOFF 5 2)
.TRAN 1U 50M UIC
.PROBE
.END
508
Fig. 11.37
Circuits and Networks
−t / 5m
Result
I(t ) = 5 + ( 2 − 5)e
5 5 2 3e2200t
PROBLEM 11.2
A series RLC circuit shown in Fig. 11.38, comprising
R 5 10 V, L 5 0.5 H, and C 5 1 F, is excited by a
constant voltage source of 100 V, using PSpice, obtain the
expression for current.
Fig. 11.38
509
Fig. 11.39
Transients
* RLC UNDERDAMPED OSCILLATIONS
V1 1
0
PWL(0 0 0.1N 100 1 100)
R
1
2
10 OHM
L
2
3
0.5 H
C
3
0
1 UF
.TRAN 10 U 50 M
.PROBE
.END
FROM EVAL GOAL FUNCTION
PEIROD(I(C)) 5 4.4715m Sec
Wd 5 1405.16 rad/sec
510
Circuits and Networks
Answers to Practice Problems
11-2.1
i(t) 5 (2  10t)e–5t
11-2.8
i(t) 5 101.2  30.9 e–0.1t – 52.11 e–4.94t
11-2.2
i = 69.7mA; VR2 = 6.97volts; VL = 3.03 volts
11-2.12
11-2.3
v(t) 5 –5e2t  15e23t volts
11-3.1
5e–5.71t
11-2.5
i1(t) 5 9.99 – 8.49 e–53104 t; i2(t)55e–5 3 104 t
11-3.4
i(t) 5 3.8  e–0.05t  0.12 e–0.31t
11-2.6
R 5 10 K ; c 5 2.5 F
11-3.8
v(t) 5 5t e–2t volts
11-2.7
i(t) 5 5 cos100t
V1(t) 5 – 4e–0.4t  4e–4999.8t
Objective-Type Questions
rrr 11.1 Transient behaviour occurs in any circuit when
(a)
(b)
(c)
(d)
there are sudden changes of applied voltage
the voltage source is shorted
the circuit is connected or disconnected from the supply
all of the above happen
rrr 11.2 The transient response occurs
(a) only in resistive circuits
(b) only in inductive circuits
(c) only in capacitive circuits
(d) both in (b) and (c)
rrr 11.3 Inductor does not allow sudden changes
(a) in currents
(c) in both (a) and (b)
(b) in voltages
(d) in none of the above
rrr 11.4 When a series RL circuit is connected to a voltage source V at t 5 0, the current passing through the inductor
L at t 5 0 is
(a)
V
R
(b) infinite
(c) zero
(d)
V
L
rrr 11.5 The time constant of a series RL circuit is
(a) LR
(b)
L
R
(c)
R
L
(d) e –R/L
rrr 11.6 A capacitor does not allow sudden changes
(a) in currents
(b) in voltages
(c) in both currents and voltages
(d) in neither of the two
rrr 11.7 When a series RC circuit is connected to a constant voltage at t 5 0, the current passing through the circuit at
t 5 0 is
(a) infinite
(b) zero
rrr 11.8 The time constant of a series RC circuit is
(a)
1
RC
(b)
R
C
(c)
V
R
(c) RC
(d)
V
vC
(d) e –RC
rrr 11.9 The transient current in a lossfree LC circuit when excited from an ac source is an ______ sine wave.
(a) undamped
(b) overdamped
(c) underdamped
(d) critically damped
511
Transients
rrr 11.10 Transient current in an RLC circuit is oscillatory when
(a) R = 2 L / C
(b) R 5 0
(c) R > 2 L / C
(d) R < 2 L / C
rrr 11.11 The initial current in the circuit shown in Fig. 11.40 when
the switch is opened for t . 0 is
(a) 1.67 A
(b) 3 A
(c) 0 A
(d) 2 A
rrr 11.12 The initial current in the circuit shown in Fig. 11.41 below
Fig. 11.40
when the switch is opened for t . 0 is
(a) 1.5 A
(b) 0 A
(c) 2 A
(d) 10 A
rrr 11.13 For the circuit shown in Fig. 11.42 the current in the 10 V
resistor when the switch is changed from 1 to 2 is
(a) 5 e20t
(b) 5 e –20t
(c) 20 e5t
(d) 20e –5t
Fig. 11.41
rrr 11.14 For the circuit shown in Fig. 11.43, the current in the 5 V resistor when the switch is changed from 1 to 2 is
1
−6
(a) 2.5e 2×10
(c) 2.5 e –10t
(b) 0
Fig. 11.42
For interactive quiz with answers,
scan the QR code given here
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(d) 5e –5t
Fig. 11.43
12
LEARNING OBJECTIVES
12.1
INTRODUCTION TO FOURIER ANALYSIS
Fourier series are series of cosine and sine terms and are used to represent general periodic signals. The ideas
and techniques of Fourier series can be extended to non-periodic signals. The signal representation for such
non-periodic signals are given by Fourier integrals and Fourier transforms.
Fourier analysis deals with Fourier series and Fourier transforms and has several applications in
mathematics, science and engineering, particularly in the area of communications and signal processing.
Any periodic signal can be represented as the sum of a finite or infinite number of sinusoidal functions, the
responses of linear systems to nonsinusoidal excitations can be determined by applying the superposition
integral. The Fourier analysis provides the ways of solving the above problem.
12.2
TRIGONOMETRIC SERIES
Periodic signals can be represented by the sum of sinusoids whose frequencies
are harmonics or integer multiples of fundamental frequency. The Fourier
series representation of a periodic signal will be of the form
x(t) 5 a0 1 a1 cos vt 1 a1 cos 2vt 1 ........... 1 b1 sin vt
1 b2 sin 2vt 1 ...
where a0, a1, a2 ........, b1, b2 .......... are real constants. Such a series is called
trigonometric series and an and bn are called the coefficients of the series.
LO 1
513
Fourier Method of Waveform Analysis
Using summation, we may write this series as
∞
x(t ) = a0 + ∑ ( an cos n vt + bn sin n vt )
n=1
Let us assume x(t) is a periodic function of period 2p and is integrable over a period 2p . Assume x(t)
v
v
can be represented by a trigonometric series.
∞
x(t ) = a0 + ∑ ( an cos n vt + bn sin n vt )
(12.1)
n=1
We assume that this series converges and has x(t) as its sum. From the function x(t), we can determine the
coefficients an and bn.
12.2.1 Determination of Constant Term a0
Integrating both sides of Eq. (12.1) from −p to p ,
v
v
p
v
we get
∫
p
v
−p
v
∞
x(t )dt = ∫ [a0 + ∑ an cos nvt + bn sin nvt ] dt
(12.2)
n=1
−p
v
The term-by-term integration gives
p
v
∫
−p
v
p
v
∞
p
v
p
v
−p
v
−p
v
x(t )dt = a0 ∫ dt + ∑ [an ∫ cos nvt dt + bn ∫ sin nvt dt ]
−p
v
n=1
The first term, on the right equals
2p
v
∫ x(t )dt =
0
(12.3)
2p
a0 . All the other integrals on the right are zero.
v
2p
a0
v
2p
v
v
a0 =
x(t )dt
2p ∫0
An alternate form of the evaluation integral with the variable vt and the corresponding period of 2p
radians is
2p
a0 =
1
x ( t ) d ( vt )
2p ∫0
(12.4)
12.2.2 Determination of the Coefficients an of the Cosine Terms
We multiply Eq. (12.1) by cos mvt, where m is any fixed positive integer and integrate from
−p p
to .
v
v
514
Circuits and Networks
p
v
p
v
∫
∞
( an cos nvt + bn sin nvt )]cos mvt dt
∫ [a0 + ∑
n=1
x(t ) cos mvt dt =
−p
v
(12.5)
−p
v
Integrating term-by-term, the right-hand side becomes,
p
v
∞
p
v
p
v
−p
v
−p
v
(12.6)
a0 ∫ cos mvt dt + ∑ [an ∫ cos nvt cos mvt dt + bn ∫ sin nvt cos mvt dt ]
n=1
−p
v
The first and third integrals are zero. The second integral becomes
p
v
∴
an =
am p
when n 5 m
v
v
x(t ) cos nv t dt
p −∫p
v
2p
=
1
x(t ) cos nv t d (vt )
p ∫0
(12.7)
12.2.3 Determination of the Coefficients bn of the Sine Terms
We multiply Eq. (12.1) by sin mvt, where m is the fixed positive integer and integrate
p
v
∫
p
v
−p
v
−p p
to
v
v
∞
x(t ) sin mvt dt = ∫ [a0 + ∑ (an cos nvt + bn sin nvt )]sin mvt dt
(12.8)
n=1
−p
v
Integrating term-by-term, the right hand-side becomes,
p
v
∞
p
v
p
v
−p
v
−p
v
a0 ∫ sin mvt dt + ∑ [an ∫ cos nvt sin mvt dt + bn ∫ sin nvt sin mvt dt ]
n=1
−p
v
The first two integrals are zero and the last integral becomes
p
v
bn =
(12.9)
bn p
when n 5 m
v
v
x(t ) sin nvt dt
p −∫p
v
or
2p
bn =
1
x(t ) sin nvt d (vt )
p ∫0
The coefficients a0, an, and bn are called the Fourier coefficients of x(t).
(12.10)
Fourier Method of Waveform Analysis
12.3
COMPACT TRIGONOMETRIC FOURIER SERIES
515
LO 1
The sine and cosine terms of the same frequency can be combined as single sine or cosine with a phase angle.
an cos nvt + bn sin nvt = cn cos( nvt + un )
or
an cos nvt + bn sin nvt = cn sin( nvt + f n )
(12.11)
where
cn = an2 + bn2
a 
 −b 
un = tan−1  n  and f n = tan−1  n 
 b 
 a 
n
n
The Fourier series in Eq. (12.1) can be expressed in the compact form
∞
x(t ) = co + ∑ cn cos( nvt + un )
(12.12)
n=1
where c0 5 a0 is the average value of x(t) and cn; un are computed from an and bn.
12.3.1 Existence of the Fourier Series, Dirichlet Conditions
Any periodic function x(t) 5 x(t 1 T ) can be expressed by a Fourier series provided the following conditions
are satisfied.
The function x(t) is absolutely integrable over one period.
∫ | x(t ) | dt < ∞
T
where T is the period.
The function x(t) has only a finite number of positive and negative maxima in one period.
The function x(t) has a finite number of discontinuities in one period.
The above conditions are known as Dirichlet conditions and hence possess a convergent Fourier series.
EXAMPLE 12.1
Find the Fourier series for the waveform shown in Fig. 12.1.
Fig. 12.1
516
Circuits and Networks
Solution The waveform equation is given by
20
vt
2p
The average value of the waveform is
v (t ) =
a0 =
2p

1
20
(vt ) 2 
vt d (vt ) =  20 ×
∫
2p 0 2p
2 

2p
= 10
0
2p
1  20 
an = ∫   vt cos nvt d (vt )
p  2p 
0
=
=
 2p
20  vt
1


+
sin
cos
v
v
n
t
n
t
 0
2p2  n
n2
20
2p2 n2
[ cos n2p − cos 0 ]
5 0 for all integer values of n
2p
bn =
=
1  20 
  vt sin nvt d (vt )
p ∫0  2p 
 2p −20
1
20  −vt


+
v
n
t
n
n
v
t
cos
sin
=
 0
pn
2p2  n
n2
Using these sine term coefficients and the average term, the series is
v(t ) = 10 −
20
20
20
sin vt −
sin 2vt − sin 3vt
2p
p
3p
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 1*
rrr 12-1.1
rrr 12-1.2
Find the trigonometric Fourier series for sawtooth wave shown in Fig. Q.1 and plot the spectrum.
Find the trigonometric Fourier series for the waveform shown in Fig. Q.2.
Fig. Q.1
*Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
Fig. Q.2
517
Fourier Method of Waveform Analysis
rrr 12-1.3 Find the trigonometric Fourier series for the waveform shown in Fig. Q.3.
rrr 12-1.4 Find the exponential Fourier series for the waveform shown is Fig. Q.4 and plot the spectrum. Concert the
coefficients obtained into the trigonometric series coefficients.
Fig. Q.3
Fig. Q.4
Frequently Asked Questions linked to LO 1*
rrr 12-1.1
rrr 12-1.2
rrr 12-1.3
rrr 12-1.4
rrr12-1.5
Find the trigonometric form of Fourier series for the V
waveform shown in Fig. Q.1.
[BPUT 2007]
p
Explain time invariant.
[GTU Dec. 2010]
0
Discuss the effect of symmetry for a periodic
function to determine the trigonometric Fourier series –v
coefficients.
[RGTU Dec. 2013]
Find the Fourier coefficients for the waveform shown
in Fig. Q.4.
Obtain trigonometric Fourier series of the signal shown in Fig. Q. 5.
3p
2p
wt
Fig. Q. 1
[RGTU Dec. 2010]
[RTU Feb. 2011]
f(t)
1
v(t)
V
–2T
–T
T
2
0
T
2
T
3T 2T
2
t
Fig. Q. 4
12.4
–3p2 –p –p2 0 p2
–1
p 3p
2
Fig. Q.5
FUNCTIONS OF ANY PERIOD 2T
The functions considered so far had period 2p. In several applications, periodic
functions will generally have other periods.
If a functions x(t) of period 2T has a Fourier serie, then this series is
∞ 
np
np 
vt + bn sin
vt 
x(t ) = a0 + ∑ an cos
(12.13)


T
T
n=1
LO 2
With the Fourier coefficients of x(t) given by
a0 =
1
2T
T
∫
x ( t ) d ( vt )
−T
T
an =
 npvt 
1
x(t ) cos 
 d (vt ) for n = 1, 2....
∫
 T 
T −T
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
t
518
Circuits and Networks
T
 npvt 
1
bn = ∫ x(t ) sin 
 d (vt ) for n = 1, 2....
 T 
T −T
EXAMPLE 12.2
Find the Fourier series of the function shown in Fig. 12.2.
0 if

v (t ) = k if

0 if
−2p < vt < −p
−p < vt < p
(12.14)
p < vt < 2p
Fig. 12.2
Solution From Eq. (12.14), the Fourier coefficients are
2p
p
a0 =
1
1
k
v ( t ) d ( vt ) =
k d ( vt ) =
∫
∫
4 p −2 p
4p −p
2
an =
1
1
npvt
npvt
v(t ) cos
d (vt ) = ∫ k cos
d ( vt )
∫
2p − 2 p
2
2 −p
2
2p
=
(12.15)
p
2k
np
sin
np
2
(12.16)
Thus, an 5 0 if n is even and
an =
2k
np
if
n = 1, 5, 9,....., an =
−2 k
np
if
n = 3, 7,11....
we also find that bn 5 0 for n 5 1,2... Hence, the result is
v (t ) =
k 2k
+
2 p


p
1
3p
1
5p
 cos vt − cos
vt + cos
vt + .....


2
3
2
5
2
(12.17)
12.4.1 Even and Odd Functions
A function x(t) is even if x(2t) 5 x(t). The waveform of such a function is symmetric with respect to the
y-axis. A function x(t) is odd if x(2t) 5 2x(t). The function cos nvt is even and sin nvt is odd. The even and
odd functions are shown in Figs 12.3 (a and b).
Fourier Method of Waveform Analysis
519
Fig. 12.3
The sum of two or more even functions is an even function and with the addition of a constant, the even
nature of the function remains. The sum of two or more odd functions is an odd function, but the addition
of a constant removes the odd nature of the function. The product of two odd functions is an even function.
The Fourier series of an even function of period 2T is a Fourier cosine series.
∞
x(t ) = a0 + ∑ an cos
n=1
np
vt
T
(12.18)
with coefficients
a0 =
an =
1
T
2
T
T
∫ x(t ) d (vt );
0
T
∫ x (t )
0
cos
npvt
d ( vt )
T
The Fourier series of an odd function of period 2T is a Fourier sine series.
∞
x(t ) = ∑ bn sin
n=1
np
( vt )
T
(12.19)
with coefficients
T
bn =
2
npvt
x(t ) sin
d ( vt )
∫
T 0
T
 T
A periodic function x(t) is said to have half-wave symmetry if x(t ) = −x t +  where T is the period. If

2
the waveform has half-wave symmetry, only odd harmonics are present in the series. This series will contain
both sine and cosine terms unless the function is also odd or even. In any case, an and bn are equal to zero for
n 5 2,4,6 … for any waveform with half-wave symmetry.
EXAMPLE 12.3
Find the trigonometric Fourier series for the triangular even waveform shown in Fig. 12.4.
520
Circuits and Networks
Fig. 12.4
Solution The waveform shown in Fig. 12.4 is an even function
since v(t) 5 v(–t)
By observation, the average value of the wave
a0 5 5
The waveform has half-wave symmetry
v(t ) = −v (t + T )
2
The equation for the waveform
v(t) 5 10 1 (10/p)vt for 2p , vt , 0
5 10 – (10/p)vt for 0 , vt , p
(12.20)
Since even waveforms have only cosine terms,
bn 5 0 for all n
0
p
10  
1 
1



an = ∫ 10 +   vt cos nvt d (vt ) + ∫
 p  
p −p 
p 0

=

10  
10 −   vt  cos nvt d (vt )

 p  
0
p
p

10 
vt
vt

+
n
t
d
t
cos
(
)
co
o
s
n
v
t
d
(
v
t
)
−
cos nvt d (vt )
v
v
∫
∫
∫

p
p
p

−p
0


−p
 1
0
1
 p 
10 
vt
vt



 
cos
n
v
t
sin
n
t
cos
n
t
sin
n
t
+
−
+
v
v
v

 2
 −p  n2
 0 
p
p
p2 



 n
10
= 2 2 {cos 0°− cos(−np) − cos np + cos 0°}
p n
=
an =
20
p2 n2
(1− cos np)
an 5 0 for n 5 2,4,6…
521
Fourier Method of Waveform Analysis
an =
40
p2 n2
for n = 1, 3, 5...
The Fourier series is
40
40
40
v(t ) = 5 + 2 cos vt +
cos 3vt +
cos 5vt + �
2
p
(3p)
(5p) 2
EXAMPLE 12.4
Find the trigonometric Fourier series for the waveform shown in Fig. 12.5.
Fig. 12.5
Solution By inspection, the average value of the waveform an 5 0. The waveform is odd and contains only
sine terms. The expression of the waveform
 20 
v(t ) =   vt
p
for − p < vt < p
(12.21)
p
bn =
=
1  20 
  vt sin nvt d (vt )
p −∫p  p 
p
−40
20  1
vt
 =

−
sin
n
t
cos
t
v
v
cos np
n
2  2

np
n
p n
 −p
(12.22)
The Fourier series is
v (t ) =
{
}
40
1
1
1
sin vt − sin 2vt + sin 3vt − sin 4vt + �
2
3
p
4
(12.23)
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 2
rrr 12-2.1
Find the trigonometric Fourier series for the waveform
shown in Fig. Q.1 and plot the line spectrum.
Fig. Q.1
522
Circuits and Networks
Find the trigonometric Fourier series for the half-wave rectified-wave shown in Fig. Q.2 and plot
the spectrum.
rrr 12-2.3 Find the exponential Fourier series for the square wave shown in Fig. Q.3 and plot the spectrum.
rrr 12-2.2
Fig. Q.3
Fig. Q.2
Frequently Asked Questions linked to LO 2
rrr 12-2.1
Find the Fourier series of the waveform shown in Fig. Q.1 and also find the line spectrum.
[RTU Feb. 2011]
f(t)
0
p
2p
3p
wt
Fig. Q.1
12.5
COMPLEX FOURIER SERIES
The Fourier series
∞
x(t ) = a0 + ∑ ( an cos nvt + bn sin nvt )
LO 3
(12.24)
n=1
can be written in the complex form
We know that
e jnvt 5 cos nvt 1 j sin nvt
e2jnvt 5 cos nvt 2 j sin nvt
By adding the Eqs (12.25) and (12.26), we get
1
cos nvt = (e jnvt + e− jnvt )
2
(12.25)
(12.26)
Fourier Method of Waveform Analysis
523
Subtracting Eq. (12.26) from Eq. (12.25) and dividing by 2j, we get
1 jnvt
sin nvt =
(e
− e− jnvt )
2j
1
1
an cos nvt + bn sin nvt = an (e jnvt + e− jnvt ) + bn (e jnvt − e− jnvt )
2
2j
1
1
= ( an − jbn ) e jnvt + ( an + jbn ) e− jnvt
2
2
Consider a0 5 c0
1
( an − jbn ) = cn
2
1
( an + jbn ) = c−n
and
2
Equation (12.24) becomes
∞
x(t ) = c0 + ∑ (cn e jnvt + c−n e− jnvt )
(12.27)
n=1
1
cn = ( an − jbn )
2
p
=
1
x(t )(cos nvt − j sin nvt )d (vt )
2p −∫p
=
1
x(t )e− jnvt d (vt )
2p −∫p
p
1
c−n = ( an + jbn )
2
p
=
1
x(t )(cos nvt + j sin nvt )d (vt )
2p −∫p
=
1
x(t )e jnvt d (vt )
2p −∫p
p
By combining the two formulas and writing cn 5 c2n, we get
x (t ) =
∞
∑
cn e jnvt
(12.28)
n=−∞
p
where
cn =
1
x(t ) e− jnvt d (vt ) for n = 0, ± 1, ± 2 �
2p −∫p
This is called complex form of Fourier series or complex Fourier series of x(t). The cn are called the
complex Fourier coefficients of x(t).
524
Circuits and Networks
EXAMPLE 12.5
Find the complex Fourier series for the square wave shown
in Fig. 12.6.
Solution The expression of the waveform
v(t) 5 210 for 2p , vt , 0
5 10 for 0 , vt , p
The average value of the wave is zero
Fig. 12.6


1 

− jnvt
d (vt ) + ∫ (10)e− jnvt d (vt )
 ∫ (−10)e

2p 

0


−p
p
0
cn =
=
0
  1
 1 − jnvt  p 
10 
 
− jnvt 


 
−
+
e
e

 − jn

 − jn
 
2p 

−
p
0

=
10 ( 0
−e + e jnp + e− jnp − e 0 )
− j 2pn
=
j10 ( jnp )
e −1
np
For n even,
e jnp 5 1
For n odd,
e jnp
(12.29)
and cn 5 0
− j ( 20)
= −1 and cn =
np
The Fourier series is
x (t ) = n + j
12.6
20 − j 3vt
20
20
20 j 3vt
e
− j e − j vt − j e j vt − j
e
p
p
3p
3p
AMPLITUDE AND PHASE SPECTRUMS
Fourier coefficient cn of the exponential form is a complex quantity and can be represented by
cn 5 Re[cn] 1 j Im[cn]
(12.30)
LO 3
(12.31)
The real part of cn is
2p
Re[cn ] =
1
x(t ) cos nvt
2p ∫0
(12.32)
The imaginary part of cn is
2p
Im[cn ] =
1
x(t ) sin nvt
2p ∫0
(12.33)
525
Fourier Method of Waveform Analysis
Re[cn] is an even function of n, where as Im [cn] is an odd function of n. Therefore, the amplitude spectrum
of the Fourier series is given by
| cn |= {Re 2 [cn ] + Im 2 [cn ]}1/ 2
(12.34)
while the phase spectrum is given by
 Im[cn ] 

un = tan−1 
 Re[cn ] 


12.7
(12.35)
THE FREQUENCY SPECTRUM
LO 3
Consider the Fourier series
∞
x(t ) = a0 + ∑ ( an cos n vt + bn sin n vt )
n=1
Let an 5 cn cos an
and bn 5 cn sin an
(12.36)
∞
Therefore, x(t ) = a0 + ∑ (cn cos a n cos nvt + cn sin a n sin nvt )
n=1
∞
= a0 + ∑ cn cos( nvt − a n )
(12.37)
n=1
From Eq. (12.36), we have
b
tan a n = n
an
b 
a n = tan−1  n 
 an 
Also, we have
(12.38)
(12.39)
cn = an2 + bn2
The magnitude of Cn is plotted against nv, the graph obtained is called the frequency spectrum of the
given waveform x(t). The amplitudes decreases rapidly for waves with rapidly convergent series. Waves with
discontinuities such as sawtooth and square waves have spectra with slowly decreasing amplitudes since their
series have strong high harmonics.
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 3
rrr 12-3.1
Find the exponential Fourier series for the fullwave rectified sine wave shown in Fig. Q.1 and
plot the spectrum.
rrr 12-3.2
Find the exponential Fourier series for the
waveform shown in Fig. Q.2 and plot the line
spectrum.
Fig. Q.1
526
rrr 12-3.3
Circuits and Networks
Find the exponential Fourier series for the waveform shown in Fig. Q.3 and plot the spectrum.
Fig. Q.2
rrr 12-3.4
rrr 12-3.5
Fig. Q.3
Find and sketch the frequency spectrum of rectangular pulse shown in Fig.Q.4.
Find and sketch the Fourier transform of the modulated signal x(t) cos 10t in which x(t) is shown
in Fig. Q.5.
Fig. Q.4
Fig. Q.5
Frequently Asked Questions linked to LO 3
rrr 12-3.1
Expand the square-wave voltage signal, as shown in the following figure into a Fourier series.
[RGTU June 2014]
f(t)
1
3T
4
–T 4
0
T4
–1
t
3T
4
Fig. Q.1
12.8
FOURIER TRANSFORM
Fourier series are powerful tools in dealing various problems involving periodic
functions. In several applications, many practical problems involve non-periodic
functions. The method of Fourier series can be extended to such non-periodic
functions.
The exponential form of Fourier series for periodic signal is given below for
convenience.
x�(t ) =
∞
∑
n=−∞
X n e jnv0t
LO 4
(12.40)
Fourier Method of Waveform Analysis
Xn =
Xn =
1
T0
1
T0
T
2
527
x�(t )e− jnv0t dt
∫
−T
2
T
2
x�(t )e− jn 2pf0t dt
∫
−T
2
where f0 is the fundamental frequency of the periodic signal x(t).
If we define non-periodic signal
x(t ) = x�(t ) for | t | <
T0
2
(12.42)
5 0 otherwise
Fig. 12.7
Xn =
1
T0
∞
T0 X n =
∫
T0
2
∫
x(t ) e− jn 2pf0t dt
(12.43)
−T0
2
x(t ) e− jn 2pf0t dt
(12.44)
−∞
We can define the envelope T0 Xn as
∞
X ( v) =
∫
x(t ) e− jvt dt
(12.45)
−∞
where v 5 2p n f0
Therefore, the coefficient Xn becomes
1
X ( v)
T0
1
Xn =
X ( nv0)
T0
Xn =
or
(12.46)
528
Circuits and Networks
Substituting Eq. (12.46) in Eq. (12.45), we get
or
∞
x�(t ) =
1

 X ( nv0 ) e jnv0t


n=−∞  T0
x�(t ) =
1
T0
∑
∞
∑
X ( nv0 ) e jnv0t
n=−∞
Substituting v0 =
2p
T0
x�(t ) =
v0 ∞
∑ X (n v0 )e jn v0t
2p n=−∞
x�(t ) =
1 ∞
∑ X (n v0 )e jn v0t v0
2p n=−∞
(12.47)
In the above complex sum, let the frequency v0 approach to zero, the index n approach to infinity such
that the product nv0 approaches a continuous frequency variable v and the discrete sum is replaced by a
continuous integral.
Therefore, x�(t ) → x(t ) as v0 approaches zero.
Then Eq. (12.45) becomes
∞
X ( v) =
∫
x(t )e− jvt dt
(12.48)
−∞
and Eq. (12.47) becomes
∞
x (t ) =
1
j vt
∫ X ( v) e d v
2p −∞
(12.49)
These expressions define a Fourier transform pair for the signal x(t) and are denoted by the notation
x(t) ↔X(v)
The integral Eq. (12.44) is referred to as Fourier transform of x(t) and is denoted by
X(v) 5 F[x(t)]
The integral Eq. (12.49) is referred to as the inverse Fourier transform and is often denoted by the symbol
x(t ) = F −1[ X (v)] .
Fourier transform is essentially a transformation from a function of the time variable t to the frequency
variable v. Not all functions can be transformed but satisfying the Dirichlet conditions in any finite interval
and
∞
∫
| x(t ) | dt < ∞
−∞
are sufficient conditions for the existence of Fourier transforms.
Fourier transform may be represented by writing X(v) in terms of magnitude and phase as
X (v) =| X (v) | e− ju( v)
(12.50)
Fourier Method of Waveform Analysis
529
Plots of |X(v)| and u(v) = X (v) versus frequency f are referred to as the amplitude and phase spectra of
x(t) respectively.
12.9
THE ENERGY SPECTRUM
LO 4
The signal is defined as an energy signal if it has finite energy over the interval (2`,`). So that power is zero.
The energy of a signal can be expressed in the frequency domain.
The normalised energy for a signal is defined
∞
∫
E=
x 2 (t ) dt
(12.51)
−∞
Using the Fourier transform representation for x(t), that is
∞
x (t ) =
1
j vt
∫ X ( v) e d v
2p −∞
(12.52)
we can write the energy as
∞
∫
E=
−∞
=
 1 ∞

j vt

 dt
(
)
(
)
x
t
X
v
e
d
v

∫  2p ∫

−∞
 −∞
∞
x 2 (t ) dt =
∞
∞

1
 x(t ) e jvt dt  d v
(
X
)
v
∫

∫
2p −∞
−∞

∞
=
1
∫ X (v) X (−v) d v
2p −∞
(12.53)
Therefore, the energy
∞
E=
1
∗
∫ X ( v) X ( v) d v
2p −∞
since
X (−v) = X ∗ (v)
∞
=
1
2
∫ | X ( v) | d v
2p −∞
∞
E=
∫
(12.54)
∞
x 2 (t ) dt =
−∞
1
2
∫ | X ( v) | d v
2p −∞
(12.55)
This is referred to as Parseval’s theorem for Fourier transforms. Thus, |X(v)|2 has the units of energy density
with frequency and is called energy density spectrum of x(t).
G (v) = | X (v) |2
Integration of G(v) over all frequencies gives the total energy contained in a signal. Similarly, integration
of G(v) over a finite range of frequencies gives the energy contained in a signal within the limits of integration.
530
Circuits and Networks
Frequently Asked Questions linked to LO 4
rrr 12-4.1
Find the exponential form of Fourier series for the waveform shown in Fig. Q.1.
[BPUT 2007]
V
p
0
3p
2p
wt
–v
Fig. Q.1
12.10
FOURIER TRANSFORM OF POWER SIGNALS
The signal is defined as power signal if and only if it has finite power and infinite
energy. The power signals are not absolutely integrable. Power signals do not satisfy
the condition.
LO 5
∞
∫
| x(t ) | dt < ∞
−∞
This class of signals are called power signals because the signal energy is infinite over the entire internal,
but the power is finite. Therefore, the power
1
P = lim
T →∞ T
∫
T
2
−T
2
(12.56)
x 2 (t ) dt < ∞
EXAMPLE 12.6
Determine the Fourier transform of a signal given in Fig.
12.8.
Solution From the waveform shown in Fig. 12.8,
−T0
T
<t < 0
2
2
= 0 otherwise
The Fourier transform of the expression
x (t ) = V
for
T0
2
X ( v) =
∫Ve
− j vt
(12.57)
Fig. 12.8
dt
−T0
2
=
T0 
V  jv T0
 e 2 − e− j v 2 

jv 
(12.58)
531
Fourier Method of Waveform Analysis
T0
 jv T0
 e 2 − e− j v 2

X (v) = VT0 
T

j 2v 0

2






 T 
sin v 0 
 2
= VT0
T
v 0
2
The Fourier transform of the signal is illustrated in Fig. 12.9.
(12.59)
Fig. 12.9
EXAMPLE 12.7
Determine the Fourier transform of the general impulse, x(t) 5 Ad(t).
Solution The impulse function for the given equation is shown in Fig. 12.10.
The transform of the above function is
∞
X (v) = ∫ A d(t ) e− jvt dt
(12.60)
−∞
X ( v) = A
The transform of the impulse function is shown in Fig. 12.11.
Fig. 12.10
Fig. 12.11
532
Circuits and Networks
EXAMPLE 12.8
Determine the Fourier transform of the signal shown in Fig.
12.12.
Solution This waveform does not satisfy the condition
∞
∫
| X (t ) | dt < ∞
(12.61)
−∞
Fig. 12.12
However, the transform of x(t) in Fig. 12.12 can be found,
if a is allowed to tend to zero. Therefore, Fig. 12.12 can be represented by Fig. 12.13.
The transform of the signal in Fig. 12.13 can be determined.
0
∫
X ( v) =
−∞
=
Fig. 12.13
∞
e at e− jvt dt + ∫ e−at e− jvt dt
0
1
1
2a
+
= 2
a − jv a + jv a + v2
(12.62)
As a → 0, X(v) is zero except when v 5 0
where X (v) = lim
a→0
2
=∞
2a
Thus, the transform is an impulse whose strength
may be obtained by integrating X(v) over the frequency
range.
∞
Therefore,
∫
∞
X ( v) d v =
−∞
Fig. 12.14
∫
−∞
2a
a + v2
2
d v = 2p
(12.63)
The transform of the signal is shown in Fig. 12.14
EXAMPLE 12.9
Determine the Fourier transform of the signum function shown is Fig. 12.15.
Fig. 12.15
Fourier Method of Waveform Analysis
533
Solution The function is defined as
sign(t) 5 1
for t . 0
5 0 for t 5 0
5 21 for t , 0
(12.64)
The Fourier transform of the function
∞
X ( v) =
∫
x(t )e− jvt dt
−∞
0
=
∫
−∞
∞
(−1)e− jvt dt + ∫ e− jvt dt
(12.65)
0
2
+1 1
+
=
jv jv
jv
The Fourier transform of signum function is shown in Fig. 12.16.
X ( v) =
(12.66)
Fig. 12.16
EXAMPLE 12.10
Determine the Fourier transform of unit step function shown in Fig. 12.17.
Fig. 12.17
Solution The function is defined as
u(t) 5 1 for t . 0
5 0 for t , 0
(12.67)
This function is obtained by the sum of two functions shown in Fig. 12.14 and in Fig. 12.16 and dividing
by 2. The transform is half the sum of transforms of individual time functions.
534
Circuits and Networks
X ( v) =

1 2
1

+ 2pd(v) =
+ pd(v)

jv
2  jv
The Fourier transform is given below.
1
X ( v) =
+ pd(v)
jv
(12.68)
(12.69)
Fig. 12.18
12.11
FOURIER TRANSFORM OF PERIODIC SIGNALS
LO 5
In the previous section, Fourier transform representation for non-periodic signals have been discussed. Fourier
transform of periodic signals have been discussed in this section. In general, Fourier transform of periodic
signals are not absolutely integrable and have infinite discontinuities. Here, we obtain the Fourier transform
of a periodic signal by Fourier transforming its complex Fourier series term-by-term. The Fourier transform
consists of a train of impulses in the frequency domain. The area of these impulses is directly proportional to
the Fourier series coefficients.
Consider a signal x(t) with Fourier transform X(v).
X (v) = 2pd(v − v0 ) ; an impulse at v 5 v0. To determine the signal x(t) for the Fourier transform X(v),
we use inverse Fourier transform relation
∞
x (t ) =
1
j vt
∫ X ( v) e d v
2p −∞
(12.70)
∞
1
=
2pd(v − v0 ) e jvt d v
∫
2p −∞
(12.71)
x (t ) = e j v0 t
Any periodic signal is represented by a linear combination of impulses equally spaced in frequency.
X ( v) =
∞
∑
2p X n d(v − nv0 )
(12.72)
n=−∞
From the Fourier inverse transform,
∞
x (t ) =
1
j vt
∫ X ( v) e d v
2p −∞
(12.73)
535
Fourier Method of Waveform Analysis
x (t ) =
x (t ) =
∞  ∞

1
 ∑ 2p X d(v − nv ) e jvt d v
n
0 
∫

2p −∞  n=−∞

∞
∑
(12.74)
X n e jnv0t
(12.75)
n=−∞
Equation (12.75) gives the exponential form of the Fourier series.
EXAMPLE 12.11
Find the Fourier transform of a periodic unit
impulse train shown in Fig. 12.19.
Solution The periodic signal is represented
by
Fig. 12.19
x (t ) =
∞
∑
d(t − nT )
(12.76)
n=−∞
By expanding x(t) in a Fourier series
x (t ) =
∞
∑
X n e jnv0t
(12.77)
n=−∞
T /2
where
Xn =
1
x(t ) e− jnv0t dt
T −∫
T /2
T /2
1
d(t ) e− jnv0t dt
T −∫
T /2
1
Xn =
T
1
Each of the Xn gives the same constant
T
∴ Fourier series representation of the unit impulse train is
1 ∞ jnv0t
f (t ) =
∑e
T n=−∞
Xn =
If we take the Fourier transform on both sides, we get
1 ∞
X (v) = { ∑ e jnv0t }
T n=−∞
2p ∞
=
∑ d(v − nv0 )
T n=−∞
2p
where v0 =
T
(12.78)
(12.79)
(12.80)
536
Circuits and Networks
Therefore, a unit impulse train in the time domain is transformed into a impulse train in the frequency
domain with an area of an each impulse as v0.
Practice Problems linked to LO 5
rrr 12-5.1
rrr 12-5.2
Find the Fourier transform of the signal shown in Fig. Q.1.
Find the Fourier transform of the signal shown in Fig. Q.2.
Fig. Q.1
Fig. Q.2
rrr12-5.3 Find the Fourier transform of the functions.
(i) v(t) 5 e2t u(t)
(ii) v(t) 5 e2|t| u(2t)
rrr 12-5.4 Find the Fourier transform of the functions given below.
(i)
x(t) 5 e2at cos bt (ii)
x(t) 5 sin (vct 1 u)
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
12.12
PROPERTIES OF THE FOURIER TRANSFORM
Properties of the Fourier transform facilitate the transformation from the time
domain to the frequency domain and vice versa.
LO 6
12.2.1 Linearity
The Fourier transform satisfies linearity and principle of superposition. Consider two signals x1(t) and x2(t).
If
x1(t) ↔ X1(v)
x2(t) ↔ X2(v)
then a1 x1 (t ) + a2 x2 (t ) ↔ a1 X 1 (v) + a2 X 2 (v)
(12.81)
Proof: F [a1 x1 (t ) + a2 x2 (t )]
= a1 F [ x1 (t )] + a2 F [ x2 (t )]
= a1 X 1 (v) + a2 X 2 (v)
Therefore, the linearity is proved.
(12.82)
12.12.2 Scaling
Consider the Fourier transform of x(t) is X(v).
If
x(t) ↔ X(v)
then for real constant a
1  v 
x( at ) ↔
X 
| a |  a 
(12.83)
Fourier Method of Waveform Analysis
537
Proof: Assume a . 0. Then Fourier transform of x(at) is
∞
∫
F {x( at )} =
x( at ) e− jvt dt
(12.84)
−∞
m 5 at
dm 5 adt
Let
Then
∞
∴
∫
F {x( at )} =
x ( m) e
 m
− jv 
 a
−∞
⋅
dm 1  v 
= X  
a
a a
(12.85)
If a , 0 then
∞
∫
F {x( at )} =
x( at ) e− jvt dt
(12.86)
−∞
Now,
2m 5 1 at
∞
∫
F {x( at )} =
x(−m) e
v
− j (−m) 
 a 

dm 
− 
a
−∞
 −m 
v
a 
∞
=
− j
−1
x(−m) e 
∫
a −∞
dm
−1
X ( v)
a
Combining these two values, we get
1
x( at ) ↔
X ( v)
|a|
=
(12.87)
(12.88)
12.12.3 Symmetry
x(t) ↔ X(v)
If
then
X(t) ↔ 2p X(2v)
∞
1
+ j vt
Proof: x(t ) =
∫ X ( v) e d v
2p −∞
(12.89)
Then replacing the dummy variable v by v9,
∞
2p x(−t ) =
∫
′
X ( v ′) e jv t d v ′
(12.90)
−∞
Now replace t by v
∞
2p x(−v) =
∫
′
X ( v ′) e jv v d v ′
(12.91)
−∞
Finally, replace v9 by t, thus
∞
2p x(−v) =
∫
X (t ) e jvt dt = F { X (t )}
−∞
X(t) ↔ 2p X(2v)
(12.92)
538
Circuits and Networks
12.12.4 Convolution
Fourier transform makes the convolution of two signals into the product of their Fourier transforms. There are
two types of convolution properties, one for the time domain and one for the frequency domain.
Time Convolution
If x1(t) ↔ X1(v)
x2(t) ↔ X2(v)
then
∞
y (t ) =
∫
x1 ( t) x2 (t − t) d t ↔ Y (v)
(12.93)
−∞
= X 1 ( v) X 2 ( v)

∞
− j vt 
 dt
e
x
(
t
)
x
(
t
−
t
)
d
t
∫ 1

2
∫
−∞

−∞
∞
Proof: F { y(t )} = Y (v) =

∞
− j vt 

x
(
t
)
∫ 1  ∫ x2 (t − t) e dt  d t
−∞

−∞
(12.94)
∞
=
(12.95)
Now, let k 5 t2t, then dk 5 dt and t 5 k 1 t
∞

 x ( k ) e− jv( k +t ) dt  d t
x
(
t
)

∫ 1  ∫ 2

−∞
−∞
∞
∴
Y ( v) =
∞
=
∫
∞
x1 ( t) e− jvt
−∞
∫
x2 ( k ) e− jv k dk
(12.96)
(12.97)
−∞
Y(v) 5 X1(v) X2(v)
(12.98)
Frequency Convolution
If
x1(t) ↔ X1(v)
x2(t) ↔ X2(v)
then
x1 (t ) x2 (t ) ↔
1
X 1 ( v) ∗ X 2 ( v)
2p
(12.99)
Proof: Considering the inverse transform of [ X1 (v) ∗ X 2 (v)] / 2p ,
we have
 X ( v) ∗ X 2 ( v) 
  1 2 ∞ jvt ∞

F −1  1
 =   ∫ e ∫ X 1 (u ) X 2 (v − u ) du ⋅ d v


2p



  2p  −∞
−∞
 1 2
=  
 2p 
∞
∫
−∞
∞
X1 (u ) ∫ X 2 (v − u ) e jvt d v ⋅ du
−∞
(12.100)
Fourier Method of Waveform Analysis
539
Let m 5 v2u then dm 5 dv and v 5 m 1 u
Thus,
∞
2 ∞

 x ( v) ∗ x 2 ( v) 
  1 
=
F −1  1
X
u
(
)
   ∫ 1 ∫ X 2 ( m)e j ( m+u )t dm ⋅ du


2
p



  2p  −∞
−∞
∞
∞
1
1
jmt
x1 (u ) e jut dt ⋅
∫
∫ x2 (m) e dm
2p −∞
2p −∞
= x1 (t ) ⋅ x2 (t )
=
Therefore, x1 (t ) ⋅ x2 (t ) ↔
1
X1 (v) ∗ X 2 (v)
2p
(12.101)
(12.102)
12.12.5 Frequency Differentiation and Integration
If x(t) ↔ X(v)
then (− jt ) x(t ) ↔
dx(v)
dv
Proof: Consider the Fourier transform x(t).
∞
X ( v) =
∫
x(t ) e− jvt dt
(12.103)
−∞
Differentiating with respect to v,
∞
dX (v)
= ∫ (− jt ) x(t )e− jvt dt
dv
−∞
Fourier transform of (2jt) x(t) is
(− jt ) x(t ) ↔
dX (v)
dv
(12.104)
dX (v)
dv
(12.105)
In general,
(− jt ) n x(t ) ↔
d n X ( v)
(12.106)
d vn
Similarly, in the frequency integration,
If x(t) ↔ X(v)
then
v
x (t )
↔ ∫ X ( v) d v
− jt
0
∞
X ( v) =
∫
(12.107)
x(t ) e− jvt dt
(12.108)
−∞
∞
v
∫
0
X ( v) d v =
∫
−∞
∞
v
x(t ) dt ∫ e− jvt d v
0
∞
1
e− j vt
= ∫ x(t ) dt
=
x(t ) e− jvt dt
∫
− jt
− jt −∞
−∞
540
Circuits and Networks
v
x (t )
↔ ∫ X ( v) d v
− jt
0
(12.109)
12.12.6 Time Shifting
If x(t) ↔ X(v)
Then x(t − t0 ) ↔ e− jvt0 X (v)
From the above, we can say that shift in time domain implies shift in phase in the transform.
Proof: F [ x(t − t0 ) ] =
∞
− j vt
∫ x (t − t0 ) e dt
(12.110)
−∞
Let t 2 t0 5 m or t 5 m 1 t0 : dt 5 dm
Therefore, F  x (t − t0 ) =
∞
∫
x (m) e
−∞
= e − j vt 0
∴
∞
∫
− jv(t0 + m)
(12.111)
dm
m
dm = e− jvt0 X (v)
x (m) e− jvm
−∞
x (t − t0 ) ↔ e− jvt0 X (v)
(12.112)
12.12.7 Frequency Shifting
If x(t) ↔ X(v)
then
x (t ) e j v0 t ↔ X ( v − v0 )
(12.113)
Frequency shifting or translation has an important significance in communication engineering. This
process is known as modulated signal.
Proof : F { x(t ) e jv0t } =
∞
=
∫
∞
∫
x(t ) e jv0t e− jvt dt
(12.114)
−∞
x(t ) e− j ( v−v0 )t dt
−∞
= X (v − v 0 )
Therefore,
x (t ) e j v0 t ↔ X ( v − v0 )
(12.115)
Table 12.1 Properties of the Fourier Transform
1.
Transformation
2.
Linearity
3.
Scaling
4.
Symmetry
x(t) ↔ X(v)
a1x1(t) 1 a2x2(t) ↔ a1X1(v) 1 a2X2(v)
x( at ) ↔
1  v 
X 
a  a 
X ( jt ) ↔ 2p X (−v)
541
Fourier Method of Waveform Analysis
5.
Time shifting
x ( t − t 0 ) ↔ e− j vt 0 X ( v )
6.
Frequency shifting
e j v0 t x ( t ) ↔ X ( v − v 0 )
7.
Time convolution
x1(t) * x2(t) ↔ X1(v) X2(v)
8.
Frequency convolution
9.
Reversal
10.
Time differentiation
11.
Time integration
x1(t ) x2 (t ) ↔
1
X 1 ( v) * X 2 ( v)
2p
x(–t) ↔ X(–v)
dn
dt n
x ( t ) ↔ ( j v) n X ( v)
t
∫
x(t )dt ↔
−∞
12.
Frequency differentiation
13.
Frequency Integration
X ( v)
+ pX (0)d(v)
jv
− jt x(t ) ↔
dX (v)
dv
v
x (t )
↔ ∫ X ( v) d v
− jt
0
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 6
rrr12-6.1
Find the Inverse Fourier transform shown in
Fig. Q.1. Use properties of Fourier transform to
help to determine the time function.
Fig. Q.1
Frequently Asked Questions linked to LO 6
rrr
12.13
APPLICATIONS IN CIRCUIT ANALYSIS
A voltage v(t) represented by Fourier series can be applied to a linear circuit to obtain LO 7
the corresponding harmonic terms of the current series. This result is obtained by
superposition principle. We consider each term of the Fourier series representing the
voltage as a single source as shown in Fig. 12.20. The equivalent impedance of the
network at each harmonic frequency is used to compute the current at that harmonic.
The sum of these individual responses is the total response i in series form due to the applied voltage.
542
Circuits and Networks
Fig. 12.20
The non-sinusoidal voltage v(t) is represented by a Fourier series
∞
v (t ) = a0 + ∑ ( an cos nvt + bn sin nvt )
(12.116)
n=1
The effective value of a voltage waveform is given by
Veff =
1
T
T
∫
2
v(t ) dt
0
From Eq. (12.116), we have

 1 T
Vrms = Veff =  ∫

T


 0
1/ 2
∞

 2 
 a + ∑ ( a cos nvt + b sin nvt ) dt 
n
n
 0




n=1

1/ 2

 2 1 2
2
2
2
2
2
=
a0 +  a1 + a2 + a3 + � + b1 + b2 + b3 + � 


2


1/ 2


1
=  A02 + ( A12 + A22 + A32 + �)


2
(12.117)
where An2 = an2 + bn2 , A0 is the average value and A1, A2, A3... are the amplitudes of the harmonics
If
v (t ) = V0 + ∑ Vn sin (nvt + f n )
and i (t ) = I 0 + ∑ I n sin (nvt + un )
Then their effective values are given by

1/ 2
1
Vrms = V02 + (V12 + V22 + V32 + �)

2


1/ 2
1
I rms =  I 02 + ( I12 + I 22 + I 32 + �)


2
or
1/ 2
2
2
2
Vrms = V02 + Vrms1
+ Vrms2
+ Vrms3
+ �
(12.118)
543
Fourier Method of Waveform Analysis
1/ 2
2
2
2
I rms =  I 02 + I rms1
+ I rms2
+ I rms3
+ �
The average power is given by
P=
=
(12.119)
T
1
T
∫ v(t ) i(t ) dt
1
T
∫ V0 + ∑Vn sin (nvt + fn )  I 0 + ∑ I n sin (nvt + un ) dt
0
T
0
After simplification, we get
P=
1
T
T
∫ V0 I 0 dt +
0
1
T
T
∫ Vn sin (nvt + fn ) I n sin (nvt + un ) dt
(12.120)
0
1
= V0 I 0 + ∑ Vn I n cos (f n − un )
2
= V0 I 0 + ∑ Vrms n I rms n cos  n
(12.121)
where n 5 fn – un
and Vrms n =
Vn
; I rms n =
In
2
2
The power factor is the ratio of average power to the apparent power.
Apparent power Vrms I rms =
The power factor p.f =
(
) (
2
2
V02 + Vrms1
+ Vrms2
+� ×
2
I 02 + I rms1
+ I r2ms2 + �
)
V0 I 0 + (Vrms n I rms n cos f n )
2
2
2
2
(V02 +Vrms1
+ Vrms2
+ �)( I 02 + I rms1
+ I rms2
+ �)
EXAMPLE 12.12
For the circuit shown in Fig. 12.21, find the current and the average power where v(t) 5 100 1 50 sin vt 1 25
sin3 vt.
Fig. 12.21
Solution At each frequency, we have to calculate the equivalent impedance of the circuit.
At v 5 0, impedance Z 5 5 V
V
100
and I 0 = 0 =
= 20 A
5
R
544
Circuits and Networks
At v 5 500 rad/s, impedance Z1 5 5 1 j (500)(0.02)
5 5 1 j 10 V
i1 (t ) =
V1
50
sin (vt − u1 ) =
sin (vt − 63.4°)
Z1
11.15
i1 (t ) = 4.48 sin (vt − 63.4°) A
At v 5 1500 rad/s, impedance Z3 5 5 1 j (1500)(0.02)
5 5 1 j 30
i3 (t ) =
V3
25
sin (3vt − u3 ) =
sin (3vt − 80.54°)
Z3
30.4
i3 (t ) = 0.823 sin (3vt − 80.54°) A
The total current
i(t) 5 20 1 4.48 sin(vt – 63.4°) 1 0.823 sin (3vt – 80.54°) A
The effective value of the current i(t)
I rms = 20 2 +
4.482 0.8232
+
= 20.25 A
2
2
The average power
2
P = I rms
R
= (410.6) 5 = 2053 W
Additional Solved Problems
PROBLEM 12.1
Find the Fourier series of the periodic function v(t) as shown in Fig. 12.22.
Fig. 12.22
Solution The equation for Fig. 12.22 is given by
v(t ) = −10 for − p < vt < 0
= 10 for 0 < vt < p
(12.122)
Fourier Method of Waveform Analysis
545
We obtain a0 5 0, since the area under the curve of v(t) between –p and p is zero.
p
an =
1
v(t ) cos nvt d (vt )
p −∫p
0
p

1 
(
)
=  ∫ (−10) cos nvt d vt + ∫ (10) cos nvt d (vt )
p −p

0

0
p
1
sin nvt 
sin nvt  
= −10
 + 10
 
p 
n −p
n 0 
5 0 because sin vnt 5 0 at –p, 0 and p for n 5 1, 2...
p
bn =
1
v(t ) sin nvt d (vt )
p −∫p
0
p

1 

 ∫ (−10) sin nvt d (vt ) + ∫ (10) sin nvt d (vt )
p −p

0

0
p
1
cos nvt 
cos nvt  
= (10)
 −10
 
p 
n −p
n 0 
=
10
[ cos 0 − cos (−np) − cos np + cos 0]
np
20 (
1− cos np)(∵ cos p = −1, cos 2p = 1, cos 3p = −1)
=
np
bn =
1 – cos np 5 2 for odd n
5 0 for even n
The Fourier series of v(t) is
v (t ) =

40 
1
1
sin vt + sin 3vt + sin 5vt + �
3
p 
5
PROBLEM 12.2
Find the trigonometric Fourier series for the waveform shown in Fig. 12.23 and plot the spectrum.
Fig. 12.23
(12.123)
546
Circuits and Networks
Solution The equation for Fig. 12.23 is given by
50
vt 0 < vt < p
p
=0
p < vt < 2p
The average value of the wave is
v (t ) =
(12.124)
p
1
50
av =
vt d ( vt )
∫
2p 0 p
=
p
 25 12.5
1  50  vt  
50  p
 − 0 =
=
  =
 4p
p
2p  p  2 0  2p2  2
Since the wave is neither even nor odd, the series contains both sine and cosine terms.
p
an =
1  50 
  vt cos nvt d (vt )
p ∫0  p 
=
p
50  1
vt


cos
n
t
+
sin
v
v
n
t
 0
n
p2  n2
=
50 [
cos np −1]
p2 n2
an = 0 for n is even (∵ cos np −1 = 0)
an =
−100
for n odd
p2 n2
p
1  50 
bn = ∫   vt sin nvt d (vt )
p p
0
=
 p −50
vt
50  1

 =
v
t
v
−
sin
cos
n
t
cos np
 0
pn
n
p2  n2
−50
pn
−50
bn =
pn
bn =
for n even
for n odd
The Fourier series of the waveform is
v(t ) = 12.5 −
+
100
p
2
cos vt −
100
(3p)
2
cos 3vt −
50
50
50
sin vt −
sin 2vt +
sin 3vt
p
2p
3p
The spectrum is given in Fig. 12.24.
100
(5p)5
cos 5vt −…
(12.125)
Fourier Method of Waveform Analysis
547
Fig. 12.24
PROBLEM 12.3
Find the exponential Fourier series for the waveform shown in Fig. 12.25 and sketch the spectrum.
Fig. 12.25
Solution The equation for the waveform shown in Fig. 12.25 is given by
100 
v(t ) = 100 + 
vt for 0 < vt < p
 p 
100 
(12.126)
v(t ) = 100 − 
vt for − p < vt < 0
 p 
The waveform is even and the coefficients of An are pure real. By inspection, the average value is
100
C0 =
= 50 V
2
p
 0 


1 
100   − jnvt
100   − jnvt


100 −


(
)
+
vt  e
d (vt )
Cn =
v
t
e
d
v
t
100
+


 ∫ 
∫




p  
p  
2p 

0 


−p 
p
p
0



100 

= 2  ∫ vt e− jnvt d (vt ) + ∫ (−vt )e− jnvt d (vt ) + ∫ pe− jnvt d (vt )

2p 

0
−p


−p
p
0




 − jnvt
 
100  e− jnvt
 − e
 
jn
t
)
(
−
jn
v
t
−
1
)
v
= 2
(
−
−
1



 
2
2

2p  (− jn)
 −p  (− jn)
 0 



=
100 
1− e jnp 
2 2 

p n
548
Circuits and Networks
Cn 5 0
Cn =
for n even
200
for n odd
p2 n2
The exponential Fourier series is
Fig. 12.26
v (t ) = � +
+
200
(−3p)
200
( p) 2
2
e j vt +
e− j 3vt +
200
(3p) 2
200
(−p)
2
e − j vt +
100
2
e jvt + ......
(12.127)
The spectrum is shown in Fig. 12.26.
PROBLEM 12.4
Find the exponential Fourier series of the triangular
waveform shown in Fig. 12.27.
Solution The voltage equation for the waveform is shown
in Fig. 12.27.
 50 
v(t ) = 25 +   vt
p
 50 
= 25 −   vt
p
for − p < vt < 0
Fig. 12.27
for 0 < vt < p
(12.128)
The coefficients of the exponential series
Cn =
Cn =
0
p

 50   − jnvt
1
1
  vt  e

+
+
d
(
t
)
25
v
 p  
2p −∫p 
2p ∫0

4 × 25
p2 n2

  
 25 −  50  vt  e− jnvt d (vt )
 p  


for n odd
Cn = 0 forr n even
PROBLEM 12.5
Find the trigonometric Fourier series for the half-wave rectified sine wave shown in Fig. 12.28 and sketch the
spectrum.
549
Fourier Method of Waveform Analysis
Fig. 12.28
The average value of the waveform
p
a0 =
1
50 [
50
p
− cos vt ]o =
50 sin vt d (vt ) =
∫
2p 0
p
2p
The series contains both sine terms and cosine terms
p
1
an = ∫ 50 sin vt cos nvt d (vt )
p 0
p
=
=
an =
50  −n sin vt sin nvt − cos nvt cos vt 


 0
p 
−n2 + 1
50
p(1− n2 )
100
(cos np + 1)
p (1− n2 )
an = 0
for n even
fo
or n odd
However, this expression is indeterminate for n 5 1 and, therefore, we must integrate seperately for a1,
p
p
a1 =
1
50 1
50 sin vt cos vt d (vt ) =
sin 2vt d (vt ) = 0
p ∫0
p ∫0 2
bn =
1
50  n sin vt cos nvt − sin nvt cos vt 

 =0
50 sin vt sin nvt d (vt ) =
∫
 0
p 
p 0
−n2 + 1
p
p
Here again, the expression is indeterminate for n 5 1, and b1 is evaluated separately.
p
b1 =
Fig. 12.29
p
1
50  vt sin 2vt 
2


50
sin
t
d
(
t
)
v
v
=
−
= 25
p ∫0
p  2
4  0
Then the Fourier series is
50  p
2
2
2

v (t ) = 
1 + sin vt − cos 2vt − cos 4vt − cos 6vt −… (12.129)


p
2
3
15
35

The spectrum is shown in Fig. 12.29.
550
Circuits and Networks
PROBLEM 12.6
Find the Fourier transform of a single triangular pulse shown in Fig. 12.30 and
draw its spectrum.
Solution The equation for the waveform is
 2 
v(t ) = 50 1− | t |
 T 
∞
v (v) =
Fig. 12.30
v (t ) e− jvt dt
∫
−∞
T
2
=
(12.130)


2
∫ 50 1− T | t | e
− j vt
dt
−T
2
T
2
= 50 ∫ e− jvt dt −
−T
2
T
2
 vT 
100
8×50
| t | e− jvt dt = 2 sin 2 

 4 
T −∫T
vT
2
 vT 
sin 2 

 4 
8×50 v T
= 2 ×
16
vT
 vT 2


 4 
 vT 
sin 2 

 4 
= 25 T
 vT 2


 4 
The spectrum is shown in Fig. 12.31.
2
Fig. 12.31
2
(12.131)
PROBLEM 12.7
Determine the Fourier transform of the pulse shown in Fig. 12.32.
Solution The equation of the waveform in Fig. 12.32 is given by
v(t ) = 100 cos vt
V (v) =
p
2
for
∫ 100 cos vt ⋅ e
−p
2
−p
p
< vt <
2
2
− j vt
d ( vt )
(12.132)
Fig. 12.32
(12.133)
551
Fourier Method of Waveform Analysis
p
2
 e jvt + e− jvt 
 e− jvt d (vt )
= 200 ∫ 


2
0
p
2
= 200 ∫
0
p
2
1 ( )
1
d vt + 200 ∫ e−2 jvt d (vt )
2
2
0
p
p/ 2
= 100 (vt )0
 e−2 jvt  2

+ 100 
 −2 j 
0
p

p  −50   − j 2 2
= 100 × + 
− e 0 
  e
2  j  


50 50
= 50p + − e− jp
j
j
= 50p − j 50 + j 50e− jp
V (v) = 50  p − j (1− e− jp )
(12.134)
PROBLEM 12.8
Obtain the Fourier transform of the function shown in Fig. 12.33.
Solution The Fourier transform of the waveform shown in Fig. 12.33 is
∞
V (v) =
∫
v(t ) e− jvt dt
Fig. 12.33
(12.135)
−∞
∞
= ∫ 100 e−10t e− jvt dt
0
∞
= ∫ 100 e−(10+ jv)t dt
0
V ( v) =
−100 [ −(10+ jv)t ]∞
e
0
10 + jv
V (v) =
100
10 + jv
(12.136)
552
Circuits and Networks
PROBLEM 12.9
Obtain the magnitude and phase spectrum of the waveform shown
in Fig. 12.34.
Solution The equation of the voltage waveform in Fig. 12.34 is
given by



10




v (t ) = 
−10



0






−T
≤t ≤0
2
T
for 0 ≤ t <
2
otherwise
for
V (v) = F [ v(t ) ] =
∞
∫
(12.137)
Fig. 12.34
v(t ) e− jvt dt
−∞
0
=
∫
v(t ) e− jvt dt +∫ v(t ) e jvt dt
0
T
2
−T
2
=
T
2
0
∫ 10 e
− j vt
−T
2
dt +∫ (−10) e− jvt dt
0
T
2
T
2
= ∫ 10 e jvt dt −∫ (+10) e− jvt dt
0
0
T
2
T
2
0
0
= 10 ∫ (e jvt − e− jvt ) dt = 20 j ∫ sin vt dt
T
 − cos vt  2
= 20 j 

V (v) =
v

 0
vT 
20 j 
1− cos

v 
2 
(12.138)
The magnitude of the spectrum is
V (v) =
vT 
20 
1− cos

v 
2 
(12.139)
553
Fourier Method of Waveform Analysis
Phase spectrum is
 Im (V (v))



f (v) = tan−1 




(
)
(
)
v
Re
V




 20 




1− cos vT 


v
2 

= tan−1 




0










 p


for v positive
+
=
 2
p


forr v negative
−


 2
The magnitude and phase spectrum are shown in Fig. (12.35).
(12.140)
Fig. 12.35
PROBLEM 12.10
Determine the Fourier transform of the sin c function shown in
Fig. 12.36.
x(t) = 50 sin c (2Tt)
Solution The transform pair is given by
t 
A x   ↔ AT sin c (vT )
T 
Applying duality and scaling property,
50 sin (2Tt ) ↔
Fig. 12.36
50  −v 
X

2T  2T 
(12.141)
Rectangular function is an even function, we get
50 sin (2Tt ) ↔
50  v 
X 
2T  2T 
The spectrum of sinc function is shown in Fig. 12.37.
Fig. 12.37
(12.142)
554
Circuits and Networks
PROBLEM 12.11
Find the Fourier transform of exponentially damped sinusoidal
waveform shown in Fig. 12.38.
Solution The equation for the damped sinusoidal waveform in
Fig. 12.38 is given by
x(t) 5 e–10t sin vct u(t)
The above equation can be written as
 e j v c t − e− j v c t 
 u (t )
x(t ) = e−10t 

2
j


From the transform pair,
e−at u (t ) ↔
(12.143)
Fig. 12.38
(12.144)
1
a + jv
(12.145)
Applying the frequency shifting property to the equation of x(t),

1 
1
1


X ( v) =
−

2 j 10 + j (v − vc ) 10 + j (v − vc ) 
vc
=
(10 + v)2 + vc2
(12.146)
PROBLEM 12.12
Find the Fourier transform of the normalised Gaussian pulse shown in
Fig. 12.39 is given by
Solution Let x(t) ↔ X(v)
Differentiating x(t) with respect to t, we get
2
dx(t )
= −2pt e−pt
dt
2
dx(t )
j
= − j 2pt e−pt = − j 2pt x(t )
dt
Taking Fourier transform both sides,
 dx(t ) 
 = F [− j 2pt x(t ) ]
jF 
 dt 
Fig. 12.39
(12.147)
(12.148)
Using differentiation property, we get
jj vX (v) = 2p
d
X ( v)
dv
(12.149)
555
Fourier Method of Waveform Analysis
1 d
−v
[ X ( v) ] =
X ( v) d v
2p
dX (v)
= −v d v
2p
dv
Integrating with respect to v on both sides,
−v2
ln X (v) =
+C
4p
where C is the constant of integration.
(12.150)
(12.151)
At v 5 0; C 5 ln{x(0)}
By definition,
∞
∫
e−pt e− jvt dt
(12.153)
∫
e−pt dt
(12.154)
X ( v) =
2
−∞
∞
X ( 0) =
(12.152)
2
−∞
The area under the curve is one X(0) 5 1
If we substitute in Eq. (12.152), we get
C50
If we substitute in Eq. (12.151), we get
ln X (v) =
−v2
X (v) = e 4 p
or
Thus
Fig. 12.40
−v2
4p
e
−pt 2
−v2
↔ e 4p
(12.155)
The spectrum of the normalized Gaussian pulse is shown in Fig.
12.40.
PROBLEM 12.13
Find the Fourier transform of the trapezoidal pulse shown in Fig. 12.41.
Solution By differentiating twice, we get
d 2 x (t )
dt 2
50
=
T
T−
2

 T
 T 
d(t + T ) − d t +  − d(t − T ) + d t − 


 2 


2

Fig. 12.41
(12.156)
556
Circuits and Networks
Using the time-shifting property, we get
T
T

 d 2 x (t ) 
jv
− jv 
 = 50  e jvT − e 2 − e− jvT + e 2 
F 


2 

 dt  T / 2 
Using time differentiation, we get
(12.157)
T
T
jv
− jv 
100  jvT
2 − e− j vT + e
2
e
−
e


T 


By simplyfying Eq. (12.158),
( j v ) X ( v) =
2

vT
 cos vT − cos
200 
2
X ( v) =

T 
−v2
or
(12.158)







vT
 cos
− cos vT 
200 

2
X (v) =


T 

v2
(12.159)
PROBLEM 12.14
Find the Fourier transform of a periodic pulse train shown in Fig. 12.42.
Solution The Fourier series representation of v(t) is given
∞
v (t ) =
∑
Vn e− jvnt ; v0 =
n=−∞
where
2p
T
T0
2
T /2
Vn =
1
1
v(t ) e− jnv0t dt =
∫
T −T / 2
T
∫ 10 e
−T0
2
T0
1
=
T
 1
2

e− jnv0t 
 − jnv0
 −T0
2
vT
vT

+ jn 0 0 
1
1  − jn 02 0
2 
= ×
e
−
e


T − jnv0 


2
=
M v0T
=
v0T0
 jn v0T0
 e 2 − e− jn 2


2j

 vT 
2
sin n 0 0 
 2 
2p
T
n
T





(12.160)
Fig. 12.42
− jnv0 t
dt
(12.161)
557
Fourier Method of Waveform Analysis
  nv0T0  
 
 sin 
  2  
Vn = 


np

for n ≠ 0
(12.162)
Therefore, the Fourier transform of the continuous function is
∞
F [ v (t ) ] =
V ( v) =
∑
Vn 2pd (v − nv0 )
∑
2pVn d (v − nv0 )
n=−∞
∞
n=−∞
 nv T 
 sin 0 0 

2  d (v − nv )
V (v) = ∑ 2p 

0


p
n

n=−∞
∞
(12.163)
PROBLEM 12.15
Find the Fourier transform of the train of unit impulses
shown in Fig. 12.43.
Solution
The periodic function is given by
∞
∑
v (t ) =
d (t − nT )
(12.164)
n=−∞
v(t) can be expanded in Fourier series
∞
v (t ) =
∑
Vn e jnv0t
Fig. 12.43
(12.165)
n=−∞
T /2
where
Vn =
1
v(t ) e− jnv0t dt
T −∫
T /2
=
1
d (t ) e− jnv0t dt
T −∫
T /2
(12.166)
1
T
(12.167)
T /2
Vn =
Therefore,
v (t ) =
1 ∞ jnv0t
∑e
T n=−∞
(12.168)
Taking Fourier transform on both sides,
1 ∞

V (v) =  ∑ e jnv0t 
 T n=−∞

=
2p ∞
2p
∑ d (v − nv0 ) ; v0 = T
T n=−∞
= v0
∞
∑ d (v − nv0 )
n=−∞
(12.169)
558
∴
Circuits and Networks
∞
∑
d (t − nT ) ↔ v0
n=−∞
∞
∑ d (v − nv0 )
(12.170)
n=−∞
The unit impulse train in the time domain has a transform of an impulse train is the frequency domain.
PROBLEM 12.16
A waveform shown in Fig. 12.44 (a) is applied to the network shown in Fig. 12.44(b). Calculate the current
through the resistor. Assume v 5 1 rad / s.
(a)
(b)
Fig. 12.44
Solution From Fig. 12.44 (a), the voltage waveform
v(t) 5 50 sin vt
5 2 50 sin vt
0 # vt # p
vt # 2p
The function v(t) is an even function and hence, bn 5 0 for all values of n.
2p
∴ an =
1
v(t ) cos n vt d (vt )
p ∫0
p
2p

1 

 ∫ 50 sin vt cos nvtd (vt ) + ∫ −50 sin vt cos nvtd (vt )
p 0

p

p
2p

50 

{sin(
n
1
)
t
sin(
n
1
)
t
}
d
(
t
)
n
t
n
t
+
−
−
−
{sin(
+
1
)
v
−
sin(
−
1
)
v
}
d
(
v
t
)
v
v
v


∫
p  ∫0

p

 cos( n + 1)vt cos( n −1)vt  p  cos( n + 1)vt cos( n −1)vt  0 
50 

 
 +
−
−


 p 



n +1
n −1
n +1
n −1
p 
0 



(12.171)

2 ∞ cos nvt 
100 1−
∑ 2 
 p n=2, 4,6 n −1 
The average current will be zero, since the capacitor does not allow dc current. Taking transform of the
given circuit, we have
where V (t ) =
−200 cos nvt
p( n2 −1)
559
Fourier Method of Waveform Analysis
From the circuit shown in Fig. 12.45, the current
v (t )
i (t ) =
j
10 −
vn
 1 
v ( t ) vn
=
tan−1 
10vn 
100v2 n2 + 1
Fig. 12.45
By substituting the value of v(t) and v 5 1 rad /s, we get
∞
i (t ) =
∑
n= 2, 4 ,6
 1 
tan−1 
10n 
(1 + 100 n )
−200 n
2
(12.172)
PROBLEM 12.17
Find the value of R if the average power dissipated in the resistor is 1000 watts if the voltage has the
following Fourier series.
v(t) = 200 sin vt 1 100 sin 3 vt 1 50 sin 5 vt
Solution The current through resistance R is
v (t ) 1
= [200 sin vt + 100 sin 3vt + 50 sin 5vt ]
R
R
The power dissipated in resistance R is
i (t ) =
=
1 
200  
100  
50 
200 ×
+
100
×
+
50
×





2 
R  
R  
R 
=
52500
1
[40000 + 10000 + 2500] =
2R
2R
PR =
∴
52500
= 1000 v
2R
R = 26 Ω
PROBLEM 12.18
Find the average power supplied to a network if the applied voltage and resulting current are given by
v(t) 5 100 + 25 sin 30t + 80 sin 60t + 40 sin 90t volts.
i(t) 5 12 sin (30t + 65°) + 20 sin (60t + 45°) + 15 sin (90t + 25°) amperes.
Solution The total average power is the sum of the harmonic power
1
P = [100 ×12 cos 65°+80 × 20 cos 45°+40 ×15 cos 25°]
2
1
= [1200 × 0.423 + 1600 × 0.707 + 600 × 0.906]
2
1
= [507.6 + 1131.2 + 543.6] = 1091.2 watts.
2
560
Circuits and Networks
PROBLEM 12.19
For the circuit shown in Fig. 12.46, find the output voltage v0(t) by
using the Fourier transform method.
Solution According to Kirchhoff’s law,
i(t) 5 i1(t) 1 i2(t)
and
Fig. 12.46
(12.173)
v0 (t) 5 i2 (t)
1
e−t u(t ) = 0 (t ) +   0 (t )
 2 
Converting Eq. (12.174) into Fourier transform, we have
1
1
= V0 ( jv) + ( jv)V0 ( jv)
jv + 1
2
 2 + jv 
1

= V0 ( jv) 
 2 
jv + 1
2
V0 ( jv) =
( jv + 1)( jv + 2)
Taking partial fractions,
2
2
V0 ( jv) =
−
jv + 1 jv + 2
By taking inverse Fourier transform, we have
v0(t) 5 2 e2tu(t) 2 2 e22t u(t)
ANSWERS TO PRACTICE PROBLEMS
12-1.1
v (t ) =

−200 
1
1
sin vt + sin 2vt + sin 3vt + �
2
p
3
12-1.3
v (t ) =
∞ 
 10  np 

5
10 
np 
1− cos  sin nvt �
+ ∑
 sin  cos nvt +







12
12 n=1 
12
np

 np
12-2.2
v (t ) =

2
100 
2
p
2

1 + cos vt + cos 2vt − cos 4vt + cos 6vt −�


2
3
15
p 
35

12-2.3
v (t ) =


20 
1 − j 5 vt 1 − j 3 vt
1
1


− e
+ e− jvt + e jvt − e+ jvt + e j 5vt − �
� + e



p
5
3
3
5





 1
� + 

12-3.2 v(t ) = V 




9p
2
+j
1  − j 3vt
1 − j 2vt  1
1 
+j
+  2 + j  e− jvt
e
 e

p
6p 
4p
2p 

1  1
1 
1 j 2vt  1
1 
+ +  2 − j  e jvt − j
e
+  2 − j  e j 3vt + �

 9p
4  p
2p 
4p
6p 
(12.174)
561
Fourier Method of Waveform Analysis
 vT 
sin 

 2 
12-3.4 V (v) = 2T
vT
2
X (v) = 2 sin c [ 2(v + 10) ] + 2 sin c [ 2(v −10) ]
12-3.5
Fig. 12.62
12-5.1
V(v) 5 2sinv 1 4sin2v
12-6.1
x (t ) =
2A
sin at cos v0 t
pt
Objective-Type Questions
rrr 12.1
In a periodic signal, the period T0 is doubled, the fundamental frequency v0 in the spectrum becomes
(a) doubled
(c) increased 4 times
(b) halved
(d) no change
rrr 12.2
Any periodic function can be expressed by a Fourier series when the function has
(a) infinite number of finite discontinuities in a period
(b) finite number of infinite discontinuities in a period
(c) finite number of finite discontinuities in a period
(d) infinite number of infinite discontinuities
rrr 12.3
A function x(t) is said to be even, if x(t) is
(a) x(2t)
(b) 2x(t)
(c) x(2t)
(d) x(t)
rrr 12.4
A function x(t) is said to be odd, if x(2t) is
(a) x(2t)
(b) x(2t)
(c) x(22t)
(d) 2x(t)
rrr 12.5
A periodic function x(t) is said to have half-wave symmetry if x(t) is

rrr 12.6
rrr 12.7
rrr 12.8
T

T

T
(a) −x t + 
(b) x t + 
(c) x t − 


 2
2
2
2at
Fourier transform for the signal e u(t) does not exist if
(a) a . 0
(b) a 5 0
(c) a 5 1
The Fourier transform
(a) satisfies linearity

(d) a , 0
(b) does not satisfy linearity
The Fourier transform exists, if the following condition is satisfied.
∞
∞
(a)
∫
−∞
f (t ) dt > K
(c)
∫
−∞
f (t ) dt =0
∞
(b)
∫
−∞
f (t ) dt < ∞
T
(d) −x t − 
 2
(d) none
562
rrr 12.9
Circuits and Networks
Fourier transform of the unit impulse d(t) is
(a)
(b) p
0
(c) 1
(d) d(v)
(c) 2p
(d)
2pd(v)
(d)
e j v0 t
rrr 12.10 What is the spectrum of a dc signal?
(a)
0
(b) 1
rrr 12.11 Inverse Fourier transform of d(v – v0) is
(a)
1 j v0 t
e
2p
(b)
1
2p
(c)
e− j v 0 t
rrr 12.12 The Fourier transform of the signal x(t) is
(a)
–X(v)
(b) X(–v)
(c) –X(–v)
(d) X(v)
(b) f1(t) f2(t)
(d) F1(v) / F2(v)
(b) F1(v) * F2(v)
(d) F1(v) / F2(v)
(c)
(d)
rrr 12.13 Time convolution property states that
(a) f1(t) * f2(t)
(c) F1(v) * F1*(v)
rrr 12.14 Frequency convolution property states that
(a) f1(t) * f2(t)
(c) F1(v) F2(v)
rrr 12.15 Fourier transform of the sgn(t) function is
(a)
2
jv
(b)
1
jv
For interactive quiz with answers,
scan the QR code given here
OR
visit
http://qrcode.flipick.com/index.php/270
jv
2jv
13
LEARNING OBJECTIVES
13.1
DEFINITION OF THE LAPLACE TRANSFORM
The Laplace transform is a powerful analytical technique that is widely
LO 1
used to study the behaviour of linear, lumped parameter circuits. Laplace
transforms are useful in engineering, particularly when the driving function
has discontinuities and appears for a short period only.
In circuit analysis, the input and output functions do not exist forever in time. For causal functions, the
function can be defined as f (t) u(t). The integral for the Laplace transform is taken with the lower limit at
t 5 0 in order to include the effect of any discontinuity at t 5 0.
Consider a function f (t) which is to be continuous and defined for values of t $ 0. The Laplace transform
is then
∞
∞
(13.1)
−st
L [ f (t )] = F ( s) = ∫ e f (t ) u(t ) dt = ∫ f (t )e−st dt
−∞
0
f (t) is a continuous function for t $ 0 multiplied by e–st which is integrated with respect to t between the
limits 0 and `. The resultant function of the variables is called Laplace transform of f (t). Laplace transform
is a function of independent variable s corresponding to the complex variable in the exponent of e–st. The
complex variable S is, in general, of the form S 5 s 1 jv and s and v being the real and imaginary parts
∞
respectively. For a function to have a Laplace transform, it must satisfy the condition
−st
∫ f (t ) e dt < ∞.
Laplace transform changes the time-domain function f (t) to the frequency-domain function F(s). Similarly,
the inverse Laplace transformation converts frequency-domain function F(s) to the time-domain function f (t)
as follows.
+j
1
L −1[ F ( s)] = f (t ) =
(13.2)
F ( s) e st ds
2j −∫j
564
Circuits and Networks
Here, the inverse transform involves a complex integration. f (t) can be represented as a weighted integral
of complex exponentials. We will denote the transform relationship between f (t) and F(s) as
L
f (t ) ←
→ F ( s)
In Eq. (13.1), if the lower limit is 0 then the transform is referred to as one-sided, or unilateral, Laplace
transform. In the two-sided, or bilateral, Laplace transform, the lower limit is –`.
In the following discussion, we divide the Laplace transforms into two types: functional transforms
and operational transforms. A functional transform is the Laplace transform of a specific function, such as
sinvt, t, e2at, and so on. An operational transform defines a general mathematical property of the Laplace
transform, such as binding the transform of the derivative of f (t). Before considering functional and
operational transforms, we used to introduce the step and impulse functions.
Frequently Asked Questions linked to LO 1
rrr
13.2
STEP FUNCTION
In switching operations, abrupt changes may occur in current and voltages. On
some functions, discontinuity may appear at the origin. We accommodate these
discontinuities mathematically by introducing the step and impulse functions.
LO 2
Figure 13.1 shows the
step function. It is zero for t < 0. It is denoted by k u(t).
Mathematically, it is defined as
k u(t) 5 0, t < 0
k u(t) 5 k, t > 0
Fig. 13.1
(13.3)
If k is 1, the function defined by Eq. (13.3) is the
unit step. The step function is not defined at t 5 0. In
situations where we need to define the transition between
0 – and 01, we assume that it is linear and that
k u(0) 5 0.5 K
(13.4)
Figure 13.2 shows the linear transition from 0 – to 01.
A discontinuity may occur at some time other than
t 5 0, for example, in sequential switching. The step
function occuring at t 5 a when a > 0 is shown in Fig.
13.3. A step occurs at t 5 a is expressed as k u(t – a).
Thus,
k u(t – a) 5 0, t < a
Fig. 13.2
k u(t – a) 5 k, t > a
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
(13.5)
565
Introduction to the Laplace Transform
If a > 0, the step occurs to the right of the origin, and if a < 0,
the step occurs to the left of the origin. Step function is 0 when
the argument t – a is negative, and it is k when the argument is
positive.
A step function equal to k for t < a is written as k u(a – t). Thus,
k u(a – t) 5 k, t < a
Fig. 13.3
Fig. 13.4
k u(a – t) 5 0, t > 0
(13.6)
The discontinuity is to the left of the origin when a < 0. A step
function k u(a – t) for a > 0 is shown in Fig. 13.4.
Step function is useful to define a finite-width pulse, by adding
two step functions. For example, the function k[u(t – 1) – u(t – 3)]
has the value k for 1 < t < 3 and the value 0 everywhere else, so it
is a finite-width pulse of height k initiated at t 5 1 and terminated
at t 5 3. Here, u(t – 1) is a function “turning on” the constant value
k at t 5 1, and the step function – u(t 5 3) as “turning off ” the
constant value k at t 5 3. We use step functions to turn on and turn
off linear functions.
EXAMPLE 13.1
Use step functions to write an expression for the function shown in
Fig. 13.5.
Solution The function shown in Fig. 13.5 is made up of linear
segments with break points at 0, 1, 3, and 4 seconds. Figure 13.5
consists of three linear segments as shown in Fig. 13.6.
(i) f1(t) 5 10t
for 0 , t , 1
(ii) f2(t) 5 –10t 1 20 for 1 , t , 3
(iii) f3(t) 5 20t – 40
for 3 , t , 4
Fig. 13.5
(13.7)
We use the step function to initiate and terminate these linear segments at the proper times.
(i) f1(t) 5 10t[u(t) – u(t – 1)], this function turns on at
t 5 0, turns off at t 5 1.
(ii) f2(t) 5 (– 10t 1 20) [u(t – 1) – u(t – 3)], this function
turns on at t 5 1, turns off at t 5 3.
(iii) f3(t) 5 (20t – 40) [u(t – 3) – u(t – 4)], this function turns
on at t 5 3, turns off at t 5 4.
The expression for f (t) is
f (t) 5 10t[u(t) – u(t – 1)] 1 (– 10t 1 20) [u(t – 1) – u(t – 3)
Fig. 13.6
1 (20t – 40) [u(t – 3) – u(t – 4)]
(13.8)
566
Circuits and Networks
EXAMPLE 13.2
Use step function to write the expression for the
following function.
Solution The function shown in Fig. 13.7 is a
combination of linear segments at break points 0, 2, 6,
8. To construct this function, we must add and subtract
linear functions of the proper slope. We use the step
function to start and terminate these linear segments at
the proper times.
Fig. 13.7
Figure 13.7 consists of the three linear segments with the following equations.
f1(t) 5 5t
for 0 , t , 2
f2(t) 5 10
for 2 , t , 6
f3(t) 5 – 5t 1 40
for 6 , t , 8
(13.9)
Using step function, the above equations can be written as
f1(t) 5 5t [u(t) – u(t – 2)]
f2(t) 5 10 [u(t – 2) – u(t – 6)]
f3(t) 5 (– 5t 1 40) [u(t – 6) – u(t – 8)]
(13.10)
The expression for f (t) is
f (t) 5 f1(t) 1 f2(t) 1 f3(t)
f (t) 5 5t [u(t) – u(t – 2)] 1 10 [u(t – 2) – u(t – 6)]
1 (– 5t 1 40) [u(t – 6) – u(t – 8)]
(13.11)
EXAMPLE 13.3
Use step function to write the expression for the following waveform.
Solution The waveform shown in Fig. 13.8 starts at t 5 0 and ends at
t 5 5 seconds. The equation for the above waveform is f (t) 5 4t. In terms
of unit function, the waveform can be expressed as
f (t) 5 4t [u(t) – u(t – 5)]
(13.12)
Fig. 13.8
EXAMPLE 13.4
Use step function to write the expression for the following
sinusoidal waveform.
Solution The sine wave shown in Fig. 13.9 originates at t 5 0
and ends at t 5 2 seconds. The wave equation
f (t) 5 10 sin vt for 0 , t , 2
In terms of unit step functions, the equation
f (t) 5 10 sin vt [u(t) – u(t – 2)]
(13.13)
Fig. 13.9
567
Introduction to the Laplace Transform
EXAMPLE 13.5
Use step function to write the expression for the function shown in Fig. 13.10.
Fig. 13.10
Solution The waveform in Fig. 13.10 consists of three linear segments. The function f (t) is defined as follows.
f1(t) 5 80t 1 120
for
–4,t,0
f2(t) 5 – 30t 1 120
f3(t) 5 30t – 360
for
for
0,t,8
8 , t , 12
(13.14)
In terms of unit step function
f1(t) 5 (80t 1 120) [u(t 1 4) – u(t)]
f2(t) 5 (– 30t 1 120) [u(t) – u(t – 8)]
f3(t) 5 (30t – 360) [u(t – 8) – u(t – 12)]
The expression for f (t) is
(13.15)
f (t) 5 f1(t) 1 f2(t) 1 f3(t)
f (t) 5 (80t 1 120) [u(t 1 4) – u(t)] 1 (– 30t 1 120) [u(t) – u(t – 8)]
1(30t – 360) [u(t – 8) – u(t – 12)]
13.3
(13.16)
IMPULSE FUNCTION
An impulse is a signal of infinite amplitude
and zero duration. In general, an impulse
signal doesn’t exist in nature, but some circuit
signals come very close to approximating
this definition. Due to switching operations,
impulsive voltages and currents occur
in circuit analysis. The impulse function
enables us to define the derivative at a
discontinuity, and thus to define the Laplace
transform of that derivative.
To define derivative of a function at a
discontinuity, consider that the function
varies linearly across the discontinuity as shown in Fig. 13.11.
LO 2
Fig. 13.11
568
Circuits and Networks
In Fig. 13.11 shown as  → 0, an abrupt
discontinuity occurs at the origin. When we
differentiate the function, the derivative between
1
– and 1 is constant at a value of
. For
2
t > , the derivative is – ae–a(t – ). The derivative
of the function shown in Fig. 13.11 is shown in
Fig. 13.12.
As  approaches zero, the value of f 9(t)
between 6 approaches infinity. At the same
Fig. 13.12
time, the duration of this large value is approaching zero. Furthermore, the area under f 9(t) between 6
remains constant as  → 0. In this example, the area is unity. As  approaches zero, we say that the function
between 6 approaches a unit impulse function; denoted d(t). Thus, the derivative of f (t) at the origin
approaches a unit impulse function as  approaches zero, or
f 9(0) → d(t) as  → 0
If the area under the impulse function curve is other than unity, the impulse function is denoted by K d(t),
where K is the area. K is often referred to as the strength of the impulse function.
Mathematically, the impulse function is defined
∞
∫
K d (t ) dt = k
(13.17)
−∞
d(t) 5 0, t  0
(13.18)
Equation (13.17) states that the area under the impulse function is
constant. This area represents the strength of the impulse. Equation
(13.18) states that the impulse is zero everywhere except at t 5 0. An
impulse that occurs at t 5 a is denoted by K d(t – a). The graphical
symbol is shown in Fig. 13.13. The impulse K d(t – a) is also shown
in Fig. 13.13.
An important property of the impulse function is the shifting property,
which is expressed as
Fig. 13.13
∞
∫
f (t ) d(t − a) dt = f ( a)
(13.19)
−∞
Equation (13.19) shows that the impulse function shifts out everything except the value of f (t) at t 5 a. The value
of d(t – a) is zero everywhere except at t 5 a, and, hence, the integral can be written
a +∈
∞
I=
∫
−∞
f (t ) d(t − a) dt =
∫
f (t ) d(t − a) dt
(13.20)
a −∈
But because f (t) is continuous at a, it takes on the value f (a) as t → a, so
a +∈
I=
∫
a−∈
a +∈
f ( a) d(t − a) dt = f ( a) ∫ d(t − a) dt = f ( a)
a−∈
(13.21)
569
Introduction to the Laplace Transform
We use the shifting property of the impulse function to find its Laplace transform.
∞
∞
0−
0−
L [d(t ) ] = ∫ d(t ) e−st dt = ∫ d(t ) dt = 1
(13.22)
Which is important Laplace transform pair that we make good use of the circuit analysis.
We can also define the derivatives of the impulse function and the Laplace transform of these derivatives.
The function illustrated in Fig. 13.14 (a) generates an impulse function as  → 0. Figure 13.14 (b) shows
the derivative of the impulse generating function, which is defined as the derivative of the impulse [d9(t)] as
Œ → 0. The derivative of the impulse function sometimes is referred to as a moment function, or unit doublet.
To find the Laplace transform of d9(t), we simply apply the defining integral to the function shown in
Fig. 13.14 (b) and, after integrating, let  → 0. Then,
 0−


 −1
 1

′
L{d (t )} = lim  ∫ 2 e−st dt + ∫  2  e−st dt 

→0 
 


0+
−

= lim
e s + e−s − 2
s2
se s − se−s
= lim
→0
2 s
2 s
s e + s 2 e−s
= lim
→0
2s
=s
→0
(a)
(13.23)
Fig. 13.14
(b)
For the nth derivative of the impulse function, we find that its Laplace transform simply is sn; that is,
{
}
L d n (t ) = s n
(13.24)
An impulse function can be thought of as a derivative of a step function, that is,
du(t )
(13.25)
dt
Figure 13.15 (a) approaches a unit step function as  → 0. The function shown in Fig. 13.15 (b), the
derivative of the function in 13.15 (b), approaches a unit impulse as  → 0.
d(t ) =
570
Circuits and Networks
(a)
(b)
Fig. 13.15
The impulse function is an extremely useful concept in circuit analysis where discontinuities occur at the
origin.
EXAMPLE 13.6
(a) Find the area under the function shown in Fig. 13.16. (b) What is the duration
of the function when  5 0? (c) What is the magnitude of f (0) when  5 0?
Solution (a) Area under the function is

A = ∫ f (t ) dt
(13.26)
−
Fig. 13.16

0
= ∫ f (t ) dt + ∫ f (t ) dt
−
0

0
 −1
t
1
1
= ∫  2 +  dt + ∫  2 t +  dt

 




−
0
0
 t2
 −1 t 2 t  
t 

=  2 +  + 
+  = 1


2
 
2


 − 
0
(13.27)
(b) As  → 0, the above function shown in Fig. 13.16 becomes an impulse function. The duration of
the function becomes zero.
(c) For an impulse function, the magnitude becomes infinite. Therefore, as  → 0, the magnitude of
the above function becomes infinite.
13.4
FUNCTIONAL TRANSFORMS
LO 2
A functional transform is simply the Laplace transform of a specified function of t. Because we are
limiting our introduction to the unilateral, or one-sided, Laplace transform, we define all functions to be
zero for t , 0–.
Unit Step Function
f (t) 5 u(t)
where u(t) 5 1 for
t0
5 0 for
t,0
(13.28)
Introduction to the Laplace Transform
571
∞
L [ f (t ) ] = ∫ f (t ) e−st dt
0
∞
= ∫ 1e−st dt =
0
L [ u (t ) ] =
e−st
−s
∞
=
0
1
s
1
s
(13.29)
Exponential Function
f (t) 5 e–at
(13.30)
∞
L(e−at ) = ∫ e−at ⋅ e−st dt
0
∞
= ∫ e−( s+a )t =
0
1
−1  −( s+a )t  ∞
e
=




0
s+a
s+a
1
L  e−at  =

 s+a
Cosine Function
(13.31)
cos vt
(13.32)
∞
L (cos vt ) = ∫ cos vt e−st dt
0
∞
∴
 e j vt + e − j vt 
 dt
= ∫ e−st 

2



0
∞
∞

1
=  ∫ e−( s− jv)t dt + ∫ e−( s+ jv)t dt 
2 0

0

∞
∞
1  e−( s− jv)t 
1  e−( s+ jv)t 
+ −
= −
2  s − jv 
2  s + jv 
0
0
1 1
1 
s
=
= 
+
2  s − jv s + jv  s 2 + v2
s
L (cos vt ) = 2
s + v2
Sine Function
sin vt
(13.34)
∞
L (sin vt ) = ∫ sin vt e−st dt
0
∞
=∫ e
0
=
(13.33)
−st
 e j vt − e − j vt 


dt
2j
∞
∞

1  −( s− jv)t
−( s + jv )t 
dt
−
e
dt
e


∫
2 j  ∫0

0

∞
∞






572
Circuits and Networks





( s− jv )t  ∞
−( s + jv )t  ∞ 


1  e
 
 + e
= 
−
 s + jv  

−
2j
s
j
v

 0 

 0




1  1
1 
v

=
−
=
2 j  s − jv s + jv  s 2 + v2
v
L (sin vt ) = 2
s + v2
∴
(13.35)
‰ Function tn where n is a positive Integer
∞
L(t n ) = ∫ t n ⋅ e−st dt
(13.36)
0
 t n e−st  ∞ ∞ e−st
 −
= 
nt n−1dt
∫

−
−
s
s

 0
0
∞
n
e−st t n−1dt
s ∫0
n
= L (t n−1 )
(13.37)
s
n −1
L (t n−1 ) =
L (t n−2 )
Similarly,
s
By taking Laplace transformations of t n – 2, t n – 3 ... and substituting in the above equation, we get
21
n n −1 n − 2
……
L (t n ) =
L (t n−m )
s s
s
ss
∠n
∠n 1
∠n
= n L (t °) = n × = n+1
s s
s
s
1
∴
L (t ) = 2
(13.38)
s
=
Hyperbolic Sine and Cosine Functions
∞
L (cosh at ) = ∫ cosh at e−st dt
(13.39)
0
∞  at
e + e−at  −st
= ∫ 
 e dt
2

0 
∞
∞
Similarly,
∞
∞
=
1
1
e−( s−a )t dt + ∫ e−( s+a )t dt
∫
2 0
2 0
=
1 1
1 1
s
+
= 2
2 s − a 2 s + a s − a2
(13.40)
∞
L (sinh at ) = ∫ sinh ( at ) e−st dt
0
(13.41)
573
Introduction to the Laplace Transform
∞  at
e − e−at  −st
= ∫ 
 e dt
2

0 
a
1
1
−
= 2
=
2( s − a) 2( s + a) s − a2
Table 13.1
(13.42)
List of Laplace transform pairs
f (t)
F(s)
Impulse
d(t)
1
Step
U(t)
1
s
t
1
Type
Ramp
s2
Exponential
e – at
Sine
sin vt
Cosine
cos vt
Hyperbolic sine
sinh at
Hyperbolic cosine
cosh at
Damped ramp
te – at
Damped sine
e – at sin vt
1
s+a
v
s 2 + v2
s
s + v2
a
2
s2 − a2
s
s2 − a2
1
( s + a) 2
v
( s + a) 2 + v 2
Damped cosine
13.5
OPERATIONAL TRANSFORMS
e – at cos vt
s+a
( s + a) 2 + v 2
LO 2
Operational transforms indicate how mathematical operations performed on either f (t) or F(s) are converted
into the opposite domain. The operations of primary interest are
(1) multiplication by a constant
(2) addition (subtraction)
(3) differentiation
(4) integration
(5) translation in the time domain
(6) translation in the frequency domain
(7) scale charging
574
Circuits and Networks
13.5.1 Multiplication by a Constant
From the defining integral, if
L [ f (t)] 5 F(s),
then L {K f (t)} 5 K F(s)
(13.43)
Consider a function f (t) multiplied by a constant K.
The Laplace transform of this function is given by
∞
L [ Kf (t )] = ∫ Kf (t ) e−st dt
0
(13.44)
∞
= K ∫ f (t ) e−st dt = KF ( s)
(13.45)
This property is called linearity property.
13.5.2 Addition (Subtraction)
Addition (subtraction) in the time domain translates into addition (subtraction) in the frequency domain.
Thus, if
L
f1 (t ) ←
→ F1 ( s) and
L
f 2 (t ) ←
→ F2 ( s), then
L [ f1(t) 6 f2(t)] 5 F1(s) 6 F2(s)
(13.46)
Consider two functions f1(t) and f2(t). The Laplace transform of the sum or difference of these two functions
is given by
∞
L { f1 (t ) ± f 2 (t )} = ∫ { f1 (t ) ± f 2 (t )} e−st dt
0
∞
∞
0
0
= ∫ f1 (t ) e−st dt ± ∫ f 2 (t ) e−st dt
5 F1(s) 6 F2(s)
∴
L { f1(t) 6 f2(t)} 5 F1(s) 6 F2(s)
(13.47)
The Laplace transform of the sum of the two or more functions is equal to the sum of transforms of the
individual function. This is called superposition property.
If we can use the linearity and superposition properties jointly, we have
L [K1 f1(t) 1 K2 f2(t)] 5 K1 L [ f1(t)] 1 K2 L [F2(t)]
5 K1 F1(s) 1 K2 F2(s)
(13.48)
Find the Laplace transform of the function.
f (t) 5 4t3 1 t2 – 6t 1 7
(13.49)
EXAMPLE 13.7
Introduction to the Laplace Transform
575
Solution L (4t3 1 t2 – 6t 1 7) 5 4L (t3) 1 L (t2) – 6L (t) 1 7L (1)
= 4×
∠3
s
4
+
∠2
−6
∠1
s
7
= 4 + 3− 2+
s
s
s
s
24
2
s
3
2
+7
1
s
6
(13.50)
EXAMPLE 13.8
Find the Laplace transform of the function.
f (t) 5 cos2 t
1 + cos 2t 
2

Solution L (cos t ) = L 


2
 cos 2t  1
1
= L   + L 
= L (1) + L (cos 2t ) ]
 2  2 [
 2 
=
1
s
2s2 + 4
+
=
2 s 2( s2 + 4) 2 s( s2 + 4)
(13.51)
(13.52)
EXAMPLE 13.9
Find the Laplace transform of the function.
f (t) 5 3t4 – 2t3 1 4e–3t – 2 sin 5t 1 3 cos 2t
(13.53)
Solution L (3t4 – 2t3 1 4e–3t – 2 sin 5t 1 3 cos 2t)
5 3 L (t4) – 2 L (t3) 1 4 L (e–3t) – 2 L (sin 5t) 1 3 L (cos 2t)
=3
=
∠4
s
72
s
5
5
−
−2
12
s
4
∠3
s
+
4
+4
s
1
5
− 2× 2
+ 3× 2
s+3
s + 25
s +4
3s
4
10
+ 2
− 2
s + 3 s + 25 s + 4
(13.54)
13.5.3 Differentiation
If a function f (t) is piecewise continuous then the Laplace transform of its derivative
 df (t ) 
 = SF ( s) − f (0)
L
 dt 
d
[ f (t )] is given by
dt
(13.55)
By definition,
∞
d

 df (t )  −st
 e dt
L  f (t )  = ∫ 
 dt

 dt 
0 
∞
= ∫ e−st d { f (t )}
0
(13.56)
576
Circuits and Networks
Integrating by parts, we get
∞
−st
= [e−st f (t )]∞
f (t ) dt
0 + ∫ se
0
= − f (0) + SF ( s)
(13.57)
Hence, we have
L [ f 9(t)] 5 SF(s) – f (0)
(13.58)
This is applicable to higher order derivatives also. The Laplace transform of the second derivative of f (t) is
d

L [ f 0(t ) ] = L  ( f 9(t ))
 dt

5 SL [ f 9(t)] – f 9(0) 5 S{SF(s) – f (0)} – f 9(0)
5 S2 F(s) – Sf (0) – f 9(0)
(13.59)
where f 9(0) is the initial value of the first derivative of f (t). We find the Laplace transform of the nth derivative
by successively applying the proceeding process, which leads to the general result.
 d n f (t ) 


−
n
n−1
n−2 dt ( 0)

L

 = S F ( s) − S f ( 0 ) − S
n


dt

 dt 

2
−
n−1
d f (0 )
d
− K − n−1 f (0− )
−S n−3
2
dt
dt
(13.60)
EXAMPLE 13.10
Using the formula for Laplace transform of derivatives, obtain the Laplace transform of (a) sin 3t (b) t3.
Solution (a)
Let f (t) 5 sin 3t
f 9(t) 5 3 cos 3t
f 0(t) 5 – 9 sin 3t
L [ f 0(t)] 5 s2[L f (t)] – sf (0) – f 9(0)
(13.61)
f (0) 5 0, f 9(0) 5 3
L [ f 0(t)] 5 L [– 9 sin 3t]
Substituting in Eq. (13.61), we get
L [– 9 sin 3t] 5 s2 L [ f (t)] – 3
L [– 9 sin 3t]– s2 [L (sin 3t)] 5 – 3
L [( s 2 + 9) sin 3t ] = 3
∴ L (sin 3t ) =
3
s +9
2
(b) Let f (t) 5 t3
Differentiating successively, we get
f 9(t) 5 3t2, f 0(t) 5 6t, f  (t) 5 6
(13.62)
(13.63)
577
Introduction to the Laplace Transform
By using the differentiation theorem, we get
L [ f (t)] 5 s3 L [ f (t)] – s2 f (0) – sf 9(0) – f 0(0)
(13.64)
Substituting all initial conditions, we get
L [ f (t)] 5 s3 L [ f (t)]
L [6] 5 s3 L [ f (t)]
6
= s3 L [ f (t )]
s
F ( s) = L[ f (t )] =
6
(13.65)
s4
13.5.4 Integration
If a function f (t) is continuous then the Laplace transform of its integral
t

1
L  ∫ f (t )dt  = F ( s)
 0
 s
By definition,
t
 ∞ t


L  ∫ f (t )dt  = ∫  ∫ f (t )dt  e−st dt
 0
 0  0

Integrating by parts, we get
 e−st t
∞ 1 ∞

=
f (t )dt  + ∫ e−st f (t ) dt
∫
−
s
s 0

 0
0
Since the first term is zero, we have
t

1
F ( s)
L  ∫ f (t )dt  = L [ f (t )] =
s
s
 0

∫ f (t ) dt
is given by
(13.66)
(13.67)
(13.68)
(13.69)
EXAMPLE 13.11
Find the Laplace transform of the ramp function r(t) 5 t.
t
Solution
We know that
∫ u (t ) = r (t ) = t
(13.70)
Integration of unit step function gives the ramp function.
t

L [r (t )] = L  ∫ u(t )dt 
 0

Using the integration theorem, we get
t
 1
1
L  ∫ u(t )dt  = L [u(t )] = 2
s
s
 0

1
since
L [u(t )] =
s
(13.71)
578
Circuits and Networks
13.5.5 Differentiation of Transforms
If the Laplace transform of the function f (t) exists then the derivative of the corresponding transform with
respect to s in the frequency domain is equal to its multiplication by t in the time domain.
−d
i.e. L [t f (t )] =
F ( s)
(13.72)
ds
By definition,
∞
d
d
F ( s) = ∫ f (t ) e−st dt
ds
ds
(13.73)
Since s and t are independent variables, and the limits 0, ` are constants not depending on s, we can
differentiate partially with respect to s within the integration and then integrate the function obtained with
respect to t.
∞
d
d
F ( s) = ∫ [ f (t ) e−st ] dt
ds
ds 0
∞
= ∫ f (t )[−te−st ] dt
0
∞
= −∫ {t f (t )} e−st dt = −L [t f (t )]
0
Hence,
−d
L [t f (t )] =
F ( s)
ds
(13.74)
EXAMPLE 13.12
Find the Laplace transform of function.
f (t) 5 t sin 2t
Solution Let f1(t) 5 sin 2t
L [ f1(t)] 5 L [sin 2t] 5 F1(s)
2
where
F1 ( s) = 2
s +4
4s
−d  2 

=+
L [t f1 (t )] = L [t sin 2t ] =
2
2


ds  s + 4 
( s + 4) 2
(13.75)
(13.76)
13.5.6 Integration of Transforms
If the Laplace transform of the function f (t) exists then the integral of corresponding transform with respect
to s in the complex frequency domain is equal to its division by t in the time domain.
∞
 f (t ) 
 = F ( s) ds
i.e. L 
 t  ∫
s
i.e.
f (t ) ↔ F ( s )
(13.77)
∞
F ( s) = L [ f (t ) ] = ∫ f (t ) e−st dt
0
(13.78)
Introduction to the Laplace Transform
Integrating both sides from s to `,
∞
∞ ∞

−st 

F
(
s
)
ds
=
∫
∫  ∫ f (t )e dt  ds
s
s 

By changing the order of integration, we get
∞
∞

= ∫ f (t )  ∫ e−st ds dt
 s

0
∞
 e−st 
 dt
= ∫ f (t ) 
 t 
0
579
(13.79)
(13.80)
∞
 f (t )  −st
 f (t ) 
 e dt = L 

=∫ 
 t 
 t 
0
∞
 f (t ) 
∫ F ( s) ds = L  t 
(13.81)
(13.82)
0
EXAMPLE 13.13
Find the Laplace transform of the function
f (t ) =
Solution
2 − 2e−2t
t
Let f1(t) 5 2 – 2e22t . Then
L [ f1(t)] 5 L [2 – 2e22t]
(13.83)
2
2
= L ( 2) − L ( 2 e − 2 t ) = −
s s+2
2s + 4 − 2s
4
=
=
s( s + 2)
s( s + 2)
 2 − 2e−2t  ∞
 = F ( s) ds
Hence L 
 ∫ 1
t

 0
∞
=∫
s
4
ds
s( s + 2)
By taking the partial fraction expansion (discussed in later section), we get
4
A
B
2
2
= +
= −
s( s + 2) s s + 2 s s + 2
∴
 2 − e −2 t  ∞
 = L [2 − 2e−2t ] ds
L 
 ∫
t


s
∞
∞
(13.84)
580
Circuits and Networks
∞
=∫
s
∞
2
2
ds − ∫
ds
+
s
s
2
s
= [2 log s − 2 log( s + 2)]∞
s

∞


 s 
1 

=  2 log
= −2 log 

 s + 2 
2

+
1

s  s

 2 − 2e − 2 t 
 = 2 log  s + 2 
L 

 s 
t


(13.85)
13.5.7 Translation in the Time Domain
If the function f (t) has the transform F(s) then the Laplace transform of f (t – a) u(t – a) is e2as F(s). By
definition,
∞
L [ f (t − a) u(t − a)] = ∫ [ f (t − a) u(t − a)] e−st dt
(13.86)
0
Since f (t – a) u(t – a) 5 0
5 f (t – a)
for t < a
for t > a
∞
∴ L [ f (t − a) u(t − a)] = ∫ f (t − a) e−st dt
(13.87)
v
Put t – a 5 t then t 1 a 5 t
dt 5 dt
Therefore, the above becomes
∞
L [ f (t − a) u(t − a)] = ∫ f (t )e−s( t+a ) d t
0
(13.88)
∞
= e−as ∫ f ( t) e−st d t = e−as F ( s)
0
∴
L [ f (t − a) u(t − a)] = e−as F ( s)
(13.89)
Translation in the time domain corresponds to multiplication by an exponential in the frequency domain.
EXAMPLE 13.14
If u(t) 5 1 for t $ 0 and u(t) 5 0 for t , 0, determine the Laplace transform of [u(t) – u(t – a)].
Solution The function f (t) 5 u(t) – u(t – a) is shown in Fig. 13.17.
L [ f (t)] 5 L [u(t) – u(t – a)]
(13.90)
5 L [u(t)] – L [u(t – a)]
Fig. 13.17
Introduction to the Laplace Transform
581
1
1 1
= − e−as = (1− e−as )
s
s s
1
L [ f (t ) = (1− e−as )
s
(13.91)
13.5.8 Translation in the Frequency Domain
If the function f (t) has the transform F(s) then the Laplace transform of e2at f (t) is F(s 1 a).
∞
By definition, F ( s) = ∫ f (t ) e−st dt
(13.92)
0
∞
and therefore, F ( s + a) = ∫ f (t ) e−( s+a )t dt
(13.93)
0
∞
= ∫ e−at f (t ) e−st dt = L [e−at f (t )]
(13.94)
0
∴
F(s 1 a) 5 L [e2at f (t)]
(13.95)
Similarly, we have
L [eat f (t)] 5 F(s – a)
(13.96)
Translation in the frequency domain corresponds to multiplication by an exponential in the time domain.
EXAMPLE 13.15
Find the Laplace transform of eat sin bt.
Solution Let f (t) 5 sin bt
(13.97)
L [ f (t )] = L [sin bt ] =
b
s + b2
2
Since L [eat f (t)] 5 F(s – a)
L [e at sin bt ] =
b
(13.98)
( s − a) + b
EXAMPLE 13.16
Find the Laplace transform of (t 1 2)2 et.
Solution Let f (t) 5 (t 1 2)2 5 t2 1 2t 1 4
L [ f (t )] = L [t 2 + 2t + 4] =
2
s
3
(13.99)
+
2
s
2
+
4
s
Since L [eat f (t)] 5 F(s – a)
L [et f (t )] =
2
( s −1)
3
+
2
( s −1)
2
+
4
s −1
(13.100)
582
Circuits and Networks
Table 13.2 List of operational transforms
Operation
f (t)
F(s)
Multiplication by a constant
K f (t)
K F(s)
Addition/Subtraction
f1(t) 6 f2(t)
F1(s) 6 F2(s)
First derivative (time)
df (t )
dt
SF(s) – f (0)
d 2 f (t )
Second derivative (time)
dt
2
n
d f (t )
nth derivative (time)
S 2 F ( s) − Sf (0) −
df (0)
dt
S n F(s) – S n – 1 f (0) – S n – 2 f 9(0) – S n – 3 f 0(0) ... f (0)n – 1
n
Operation
dt
f (t)
F(s)
f (t ) dt
F ( s)
s
t
∫
Time integral
0
Translation in time
Translation in frequency
Scale changing
f (t – a) u(t – a), a > 0
e2as F(s)
e2at f (t)
F(s 1 a)
f (at), a > 0
 s
F  
a  a 
First derivative (s)
t f (t)
tn f (t)
nth derivative (s)
f (t )
t
S integral
−
dF ( s)
ds
(−1) n
d n F ( s)
ds n
∞
∫ F (u) du
s
13.5.9 Scale Changing
The scale-change property gives the relationship between f (t) and F(s) when the time variable is multiplied
by a positive constant.
1 s
(13.101)
L { f ( at )} = F   , a > 0
a  a 
By definition,
∞
L [ f ( at )] = ∫ f ( at ) e−st dt
Put
at 5 t
dt =
a
dt
(13.102)
583
Introduction to the Laplace Transform
∞
L [ f ( at )] = ∫ f ( t)e
0
∞
s
− t
a
1
⋅ dt
a
s
− t
1
= ∫ f ( t )e a d t
a 0
=
1  s 
F 
a  a 
(13.103)
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 2*
Use step functions to write the expression for the function shown in Fig. Q.1.
rrr13-2.2 Step functions can be used to define a window
function. Thus, u(t – 1) – u(t – 4) defines a
window 1 unit high and 3 units wide located
on the time axis between 1 and 4.
A function f (t) is defined as follows:
f (t) 5 0, t # 0
5 30t, 0 # t # 2s
5 60, 2s # t # 4s


= 60, cos  t −  , 4 s ≤ t ≤ 8s
 4

Fig. Q.1
5 30t – 300, 8s # t # 10s
5 0, 10s # t # `
Sketch f (t) over the internal – 2s # t # 12s.
rrr 13-2.3 Evaluate the following integrals
rrr 13-2.1
3
(a) I = ∫ (t 3 + 2) [d(t ) + 8 d(t −1)] dt
−1
2
( b) I = ∫ t 2 [d(t ) + d(t + 1.5) + d(t − 3)] dt
−1
Explain why the following function generates an impulse function as  → 0
/
f (t ) =
,
−∞ ≤ t ≤ ∞
 +t
rrr 13-2.5 Find the Laplace transform of the signal t(0.5)t u(t).
rrr 13-2.6 Make a sketch of f (t) for – 25s # t # 25s when f (t) is given by the following expression:
f (t) 5 – (20t 1 400) u(t 1 20) 1 (40t 1 400) u(t 1 10)
1 (400 – 40t) u(t – 10) 1 (20t – 400) u(t – 20)
rrr 13-2.7 Find the Laplace transform of each of the following functions:
(a) te–at
(b) sin vt
(c) sin (vt 1 u)
(d) cosht
(e) cosh (t 1 u)
rrr 13-2.4
*Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
584
Circuits and Networks
Use the appropriate operational transform to find the Laplace transform of each function
d −at
(a) t2 e–at
(b)
(c) t cos vt
(e sinh bt )
dt
rrr 13-2.9 Find the Laplace transforms for (a) and (b):
d
(a) f (t ) = (e−at sin vt )
dt
rrr 13-2.8
t
f (t ) = ∫ e−ax cos vx dx
(b)
0−
(c) Verify the results obtained in (a) and (b) by first carrying out the mathematical operation
and the finding the Laplace transform.
1 s
rrr 13-2.10 ( a) Show that L [ f ( at )] = F  
a  a 
 f (t ) 
 is Laplace transformable then
(b) Show that if F ( s) = L [ f (t )] and 
 t 
∞

 f (t ) 

∫ F (u) du = L  t 
s
Hint: Use the determing integral to write
∞ ∞


−ut 
F
(
u
)
du
=
∫
∫ ∫ f (t ) e dt  du

s
s 0
and then reverse the order of integration.
rrr 13-2.11 Find the signal y(t), the Laplace transform of signal which is
∞
Y ( s) =
rrr 13-2.12
s3 + 7 s 2 + 18s + 20
s 2 + 5s + 6
Find f (t ) if F ( s) =
rrr 13-2.13 Find
10( s 2 + 119)
( s + 5) ( s 2 + 10 s + 169)
f (t) for each of the following functions:
(a) F ( s) =
(c) F ( s) =
18s 2 + 66 s + 54
( s + 1) ( s + 2) ( s + 3)
( b) F ( s) =
11s 2 + 172 s + 700
( s + 2) ( s 2 + 12 s + 100)
56 s 2 + 112 s + 5000
s( s 2 + 14 s + 625)
rrr 13-2.14 Find
f (t) of the following functions:
40
(a) F ( s) = 2
( s + 4 s + 5) 2
(c) F ( s) =
2 s3 + 8s 2 + 2 s − 4
s 2 + 5s + 4
(b) F ( s) =
5s 2 + 29 s + 32
( s + 2) ( s + 4)
585
Introduction to the Laplace Transform
Frequently Asked Questions linked to LO 2
r 13-2.1
What are the Laplace transforms of the following voltage
waveform shown in Fig. Q.1?
[BPTU 2008]
r 13-2.2
VO
a
a
Hs
rr
ht
Hs
O
t
T
Fig. Q.1
s + s+
r
r
r
t=0
i
rr
i
336 V
10 H
i1
i2
42 W
48 W
Fig. Q.7
r 13-2.8
13.6
8.4 H
LAPLACE TRANSFORM OF PERIODIC FUNCTIONS
Periodic functions appear in many practical problems. Let the function f (t) be a
periodic function which satisfies the condition f (t) 5 f (t 1 T ) for all t > 0 where
T is the period of the function.
T
2T
0
T
LO 3
L [ f (t )] = ∫ f (t ) e−st dt + ∫ f (t ) e−st dt + …
( n+1)T
+
f (t ) e−st dt + …
∫
(13.104)
nT
T
T
= ∫ f (t ) e−st dt + ∫ f (t ) e−st e−sT dt + …
0
0
T
+ ∫ f (t ) e−st e−nsT dt + …
0
T
= (1 + e−sT + e−2 sT + … + e−nsT + …) ∫ f (t ) e−st dt
(13.105)
0
=
1
1− e
−sT
T
∫
0
f (t ) e−st dt
(13.106)
586
Circuits and Networks
EXAMPLE 13.17
Find the transform of the waveform shown in
Fig. 13.18.
Solution Here,. the period is 2T
L [ f (t )] =
=
=
=
1
1− e
−2 sT
1
1− e−2 sT
1
1− e−2 sT
1
1− e−2 sT
1−
−e
=
∴
1
1− e−2 sT
2T
∫
f (t ) e−st dt
0
2T
T

 Ae−st dt + (− A)e−st dt 
∫

∫
 0

T
T
2
T
 A

− e−st + A e−st 
 s

s

0
T 
 A −sT

A
− (e
−1) + (e−2 sT − e−sT )
 s

s

A
 A 1− e−sT 

 (1− e−sT ) 2  = 
 s 1 + e−sT 
 s
L [ f (t )] =
Fig. 13.18

A 1− e−sT 


s 1 + e−sT 
(13.108)
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 3
rrr 13-3.1
(13.107)
Find the Laplace transform of the waveform shown in Fig. Q.1.
Fig. Q.1
587
Introduction to the Laplace Transform
rr 13-3.2
Find the Laplace transform of the periodic waveform shown in Fig. Q.2.
Fig. Q.2
Frequently Asked Questions linked to LO 3
ut
r
f t
r
t
f t
te
r
5W
K
1H
rr
k
t
s
r
V(t) = 6e
–2t
r
i
0.25F
Fig. Q.4
r
t
e
t
t
t i
di
dt
rr
d 2i
dt 2
di
dt
13.7
+
di
=t
dt
i
t
t i
di
dt
INVERSE TRANSFORMS
So far, we have discussed Laplace transform of a functions f (t). If the function
is a rational function of s, which can be expressed in the form of a ratio of two
polynomials in s such that no non-integral powers of s appear in the polynomials.
In fact, for linear, lumped-parameter circuits whose component values are constant,
the s-domain expressions for the unknown voltages and currents are always rational
functions of s. If we can inverse transform rational functions of s, we can solve for
the time domain expressions for the voltages and currents.
In general, we need to find the inverse transform of a function that has the form.
a s n + an−1s n−1 + … + a1s + a0
N ( s)
F ( s) =
= nm
D( s) bm s + bm−1s m−1 + … + b1s + b0
LO 4
(13.109)
588
Circuits and Networks
The coefficients a and b are real constants, and the exponents m and n are positive integers. The ratio
N ( s)
is called a proper rational function if m > n, and an improper rational function if m # n. Only a proper
D ( s)
rational function can be expanded as a sum of partial fractions.
13.7.1 Partial Fraction Expansion: Proper Rational Functions
A proper rational function is expanded into a sum of partial fractions by writing a term or a series of terms for
each root of D(s). Thus, D(s) must be in factored form before we can make a partial fraction expansion. The
roots of D(s) are either (1) real and distinct, (2) complex and distinct, (3) real and repeated, or (H ) complex
and repeated.
When the Roots are Real and Distinct
N ( s)
(13.110)
D ( s)
where D(s) 5 (s – a) (s – b) (s – c)
(13.111)
Expanding F(s) into partial fractions, we get
A
B
C
F ( s) =
+
+
(13.112)
s−a s−b s−c
To obtain the constant A, multiplying Eq. (13.112) with (s – a) and putting s 5 a, we get
In this case,
F ( s) =
F(s) (s – a)|s 5 a 5 A
Similarly, we can get the other constants.
B 5 (s – b) F(s)|s 5 b
C 5 (s – c) F(s)|s 5 c
EXAMPLE 13.18
Determine the partial fraction expansion for
2
F (s ) =
Solution
F ( s) =
s + s +1
s(s + 5)(s + 3)
s2 + s +1
s( s + 5)( s + 3)
(13.113)
s2 + s +1
A
B
C
= +
+
s( s + 5)( s + 3) s s + 5 s + 3
A = sF ( s) s=0 =
(13.114)
1
s2 + s + 1
=
( s + 5)( s + 3)
15
s= 0
B = ( s + 5) F ( s) s=−5 =
C = ( s + 3) F ( s) s=−3 =
s2 + s + 1
s( s + 3)
= 2.1
s=−5
s2 + s + 1
= −1.17
s( s + 5) s=−3
589
Introduction to the Laplace Transform
s2 + s +1
1
2.1 1.17
=
+
−
s( s + 5)( s + 3) 15s s + 5 s + 3
(13.115)
When Roots are Real and Repeated
In this case, F ( s) =
N ( s)
D ( s)
where D(s) 5 (s – a)n D1(s)
The partial fraction expansion of F(s) is
F ( s) =
A0
( s − a)
+
n
A1
( s − a)
n−1
+…+
An−1 N1 ( s)
+
s − a D1 ( s)
(13.116)
where N ( s) represents the remainder terms of expansion.
D ( s)
To obtain the constant A0, A1, ... An – 1, let us multiply both sides of Eq. (13.116) by (s – a)n.
Thus,
(s – a)n F(s) 5 F1(s) 5 A0 1 A1 (s – a) 1 A2 (s – a)2 1 ...
1 An – 1 (s – a)n – 1 1 R(s) (s – a)n
(13.117)
where R(s) indicates the remainder terms
Putting s 5 a, we get
A0 5 (s – a)n F(s)|s 5 a
Differentiating Eq. (13.117) with respect to s, and putting s 5 a, we get
d
A1 = F1 ( s) s=a
ds
1 d2
Similarly, A2 =
F1 ( s) s=a
2 ! ds2
d n F ( s)
An =
In general,
n!
ds n
(13.118)
s= a
EXAMPLE 13.19
Determine the partial fraction expansion for
F (s ) =
Solution
F ( s) =
s −5
s ( s + 2)
2
=
s −5
s(s + 2)
2
.
B
A
B
+
+ 1
2
s+2
s ( s + 2)
A = F ( s ) S s= 0 =
s −5
( s + 2)
−
2
s= 0
5
= − = −1.25
4
590
Circuits and Networks
s −5
s
s −5
B0 = F ( s) ( s + 2) 2 s=2 =
= 3.5
s s=−2
N1 ( s) = ( s + 2) 2 F ( s) =
d
F1 ( s) s=−2
ds
5
d  5
= 1− 
=
ds  3  s=−2 s 2
B1 =
=
s=−2
5
= 1.25
4
When Roots are Distinct Complex Roots of D(s)
Consider a function
F ( s) =
N ( s)
D( s)( s − a + jb)( s − a − jb)
(13.119)
The partial fraction expansion of F(s) is
F ( s) =
N ( s)
A
B
+
+
s − a − jb s − a + jb D ( s)
(13.120)
where N ( s) is the remainder term.
D ( s)
Multiplying Eq. (13.120) by (s – a – jb) and putting
S 5 a 1 jb,
we get
A=
N (a + jb)
D1 (a + jb)(+2 jb)
(13.121)
Similarly
B=
N (a − jb)
(−2 jb) D1 (a − jb)
(13.122)
In general, B 5 A* where A* is complex conjugate of A.
If we denote the inverse transform of the complex conjugate terms as f (t)


A
B

f ( t ) = L −1 
+
 s − a − jb s − a + jb 


*


A
A
−1

= L 
+

 s − a − jb s − a + jb 
(13.123)
where A and A* are conjugate terms.
If we denote A 5 C 1 jD, then
B 5 C – jD 5 A*
∴
f (t) 5 eat (Ae jbt 1 A* e2jbt )
(13.124)
Introduction to the Laplace Transform
591
EXAMPLE 13.20
Find the inverse transform of the function.
F (s ) =
Solution
F ( s) =
s +5
2
s(s + 2s + 5)
s+5
(13.125)
s( s + 2 s + 5)
2
By taking partial fractions, we have
s+5
A
B
B*
+
+
s s + 1− j 2 s + 1 + j 2
s( s2 + 2 s + 5)
s+5
A = F ( s ) s s= 0 = 2
=1
s + 2s + 5
s+5
B = F ( s)( s + 1− j 2) s=−1+ j 2 =
s( s + 1 + j 2) s=−1+ j 2
F ( s) =
=
(13.126)
2+ j
2+ j
4 + j2
−1
=
=
=
(−1 + j 2) j 4 2 j (−1 + j 2) −2 j − 4
2
s+5
B* = F ( s) ( s + 1 + j 2) s=−1− j 2 =
s( s + 1− j 2) s=−1− j 2
=
∴
=
−1 − j 2 + 5
(−1− j 2)(−1− j 2 + 1− j 2)
=
−1
4 − j2
4 − j2
2(2 − j )
=
=
=
2
−(1 + j 2)( j 4) 4 j − 8 −4(2 − j )
1
1
1
F ( s) = −
−
s 2( s + 1− j 2) 2( s + 1 + j 2)
(13.127)
The inverse transform of F(s) is f (t)
1

1
1

f (t ) = L −1[ F ( s)] = L −1 −
−
 s 2( s + 1− j 2) 2( s + 1 + j 2) 




 1 −1 
1 1
1
1

− L 
= L −1   − L −1 




 s  2
 s +1+ j 2 
 ( s + 1− j 2)  2
1
1
= 1− e(−1+ j 2)t − e(−1− j 2)t
2
2
(13.128)
When Roots are Repeated and Complex of D(S)
The complex roots always appear in conjugate pairs and that the coefficients associated with a conjugate pair are also conjugate, so that only half the
Ks need to be evolved.
Consider the function F ( s) =
768
( s + 6 s + 2 s) 2
2
(13.129)
592
Circuits and Networks
By factoring the denominator polynomial, we have
768
F ( s) =
( s + 3 − j 4) 2 ( s + 3 + j 4) 2
=
K1
( s + 3 − j 4)
+
2
+
K2
s + 3− j4
K1*
( s + 3 + j 4) 2
+
K 2*
s + 3 + j4
(13.130)
Now we need to evaluate only K1 and K2, because K*1 and K*2 are conjugate values.
The value of K1, is
768
768
K1 =
=
= −12
2 s=−3+ j 4
( s + 3 + j 4)
( j8) 2
The value of K2 is
K2 =
(13.131)

768
d 

ds  ( s + 3 + j 4) 2  s=−3+ j 4
=−
2(768)
( s + 3 + j 4 )3
s=−3+ j 4
=−
2(768)
( j8)3
= − j 3 = 3 ∠− 90°
(13.132)
From Eqs (13.131) and (13.132)
K1* 5 – 12, K2* 5 j3 5 3 ∠ 90º
We now group the partial fraction expansion by conjugate terms to obtain


−12
−12

F ( s) = 
+
2
( s + 3 + j 4) 2 
 ( s + 3 − j 4)
 3∠− 90°
3∠90° 

+ 
+
 s + 3 − j 4 s + 3 + j 4 
(13.133)
(13.134)
Inverse transform of the above function is
f (t) 5 [–24t e–3tcos 4t 1 6e–3tcos (4t – 90º)] u(t)
(13.135)
Table 13.3 Useful transform pairs
Nature of roots
Distinct real
Repeated complex
f (t)
k
s+a
Ke–at u(t)
k
Repeated real
Distinct complex
F(s)
Kt e–at u(t)
( s + a)2
k
k
+
s + a − jb s + a + jb
k
( s + a − jb)
2
+
k*
( s + a + jb)2
2|k|e–at cos (bt 1 u) u(t)
2t|k|e–at cos (bt 1 u) u(t)
593
Introduction to the Laplace Transform
13.7.2 Partial Fraction Expansion: Improper Rational Function
An improper rational function can always be expanded into a polynomial plus a proper rational function. The
polynomial is then inverse-transformed into impulse functions and derivatives of impulse functions.
Consider a function
F ( s) =
s 4 + 13s3 + 66 s 2 + 200 s + 300
s 2 + 9 s + 20
(13.136)
Dividing the denominator into the numerator until the remainder is a proper rational function gives
F ( s) = s 2 + 4 s + 10 +
30 s + 100
s + 9 s + 20
2
(13.137)
Now we expand the proper rational function into a sum of partial fractions.
30 s + 100
s + 9 s + 20
2
=
30 s + 100
50
−20
=
+
( s + 4) ( s + 5) s + 4 s + 5
(13.138)
Substituting Eq. (13.138) into Eq. (13.137) yields
F ( s) = s 2 + 4 s + 10 −
20
50
+
s+4 s+5
(13.139)
By taking inverse transform, we get
d 2 d(t )
d d(t )
+ 10 d(t ) − ( 20e−4t − 50e−5t ) u(t )
(13.140)
dt
dt
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
f (t ) =
2
+4
Practice Problems linked to LO 4
∞
rrr 13-4.1
Find f (t) if f (t ) =
F ( v) =
rrr 13-4.2
4 + jv
 d ( v)
9 + jv
Find the inverse transforms of the following functions
1

(a) 2
(b)
s+
s +9
8
5
(c)
(d) 2
( s + 3) ( s + 5)
s +9
(e)
rrr 13-4.3
1
jtw
∫ F ( v) e d v
2 −∞
K1 K 2 K3
+ 2 + 3
s
s
s
Find the inverse Laplace transform of
1
F ( s) =
( s + 2) 2
594
Circuits and Networks
13.8
INITIAL AND FINAL VALUE THEOREMS
The initial- and final-value theorems are useful because they enable us to
determine from F(s) the behaviour of f (t) at 0 and `. Hence, we can check the
initial and final values of f (t) to see if they conform with known circuit behaviour,
before actually finding the inverse transform of F(s).
The initial-value theorem states that
lim f (t ) = lim SF ( s)
t →0
LO 5
s→∞
and the final-value theorem states that
lim f (t ) = lim SF ( s)
t →∞
s→ 0
(13.141)
(13.142)
The initial-value theorem is based on the assumption that f (t) contains no impulse functions.
To prove the initial-value theorem, we start with the operational transform of the first derivative.
 df 
L   = SF ( s) − f (0)
 dt 
∞
df −st
=∫
e dt
dt
0
(13.143)
Now, we take the limit as s → `
∞
lim [SF ( s) − f ( )] = lim
s→∞
s→∞
∫
df −st
e dt
dt
(13.144)
The right-hand side of the above equation becomes zero as s → `
∴
lim [SF ( s) − f (0)] = 0
s→∞
lim SF ( s) = f (0) = lim f (t )
s→∞
t →0
(13.145)
The proof of the final-value theorem also starts with Eq. (13.143). Here, we take the limit as s → 0.
 ∞ df


lim [SF ( s) − f (0)] = lim  ∫
e−st dt 
s→ 0 

s→∞
 0 dt
(13.146)
lim [SF ( s) − f (0)] = [ f (t )]∞
0
(13.147)
s→ 0
lim SF ( s) − f (0) = lim f (t ) − f (0))
s→ 0
t →∞
Since f (0) is not a function of s, it gets cancelled from both sides.
∴
lim f (t ) = lim SF ( s)
t →∞
(13.148)
s→
The final-value theorem is useful only if f (`) exists.
The Application of Initial- and Final-Value Theorems
Consider the transform pair given by
100( s + 3)
( s + 6)( s + 6 s + 25)
2
↔ [−12e−6t + 20e−3t cos ( 4t − 53.13°)] u(t )
(13.149)
Introduction to the Laplace Transform
595
The initial-value theorem gives
 3
100 s 2 1 + 
 s 
lim [SF ( s) = lim
=0
s→∞ 3 
6   6 25 
s→∞
s 1 +  1 + + 2 
s  
s s 

lim f (t ) = [−12 + 20 cos (−53.13°)](1) = −12 + 12 = 0
t →0
(13.150)
(13.151)
The final-value theorem gives
100 s( s + 3)
lim SF ( s) = lim
s→ 0 ( s + 6 )
s→ 0
( s 2 + 6 s + 25)
=0
lim f (t ) = lim [−12e−6t + 20e−3t cos ( 4t − 53.13°)] u(t ) = 0
t →∞
t →∞
(13.152)
(13.153)
EXAMPLE 13.21
Verify the initial-value theorem for the following functions.
(a) 5e–4t
(b) 2 – e5t
Solution (a) Let f (t) 5 5e–4t
5
then
F ( s) =
s+4
5s
SF ( s) =
s+4
5
lim SF ( s) = lim
=5
s→∞
s→∞ 1 + 4
s
lim f (t ) = lim 5e−4t = 5
t →0
t →0
(13.154)
(13.155)
Hence, the initial-value theorem is proved.
(b) Let f (t) 5 2 – e5t
Then
(13.156)
F(s) 5 L (2 – e5t) 5 L (2) – L (e5t)
2
1
s −10
= −
=
s s − 5 s ( s − 5)
s −10
SF ( s) =
s −5
 10 
1− 

s 
lim SF ( s) =
=1
 5 
s→∞
1− 
 s 
lim f (t ) = lim (2 − e5t ) = 1
t →0
t →0
Hence, the initial-value theorem is proved.
(13.157)
596
Circuits and Networks
EXAMPLE 13.22
Verify the final-value theorem for the following functions.
(a) 2 1 e–3t cos 2t
(b) 6(1 – e–t)
–3
t
Solution (a) Let f (t) 5 2 1 e cos 2t
2
s+3
then
F ( s) = +
s ( s + 3) 2 + 4
SF ( s) = 2 +
s2
+
3s
( s + 3) + 4 ( s + 3) 2 + 4

s( s + 3) 
lim SF ( s) = lim  2 +
=2
s→ 0
s→ 0 
( s + 3) 2 + 4 

lim f (t ) = lim [2 + e−3t cos 2t ] = 2
t →∞
2
t →∞
Hence, the final-value theorem is proved.
(b) Let f (t) 5 6(1 – e–t)
then
(13.158)
(13.159)
(13.160)
6
6
6
F ( s) = −
=
s s + 1 s( s + 1)
6
SF ( s) =
s +1
lim SF ( s) = 6
s→ 0
lim f (t ) = lim 6(1− e−t ) = 6
t →∞
t →∞
(13.161)
Hence, the final value theorem is proved.
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 5
Apply the initial- and final-value theorems to each transform pair in Problem 13.2.13 (refer
page 584).
rrr 13-5.2 Use the initial- and final-value theorems to find the initial- and final-values of f (t) for the
following functions.
4 s2 + 7s + 1
7 s 2 + 63s + 134
(a) F ( s) =
(b) F ( s) =
( s + 3) ( s + 4) ( s + 5)
s( s + 1) 2
40
(c) F ( s) = 2
( s + 4 s + 5) 2
rrr 13-5.1
rrr 13-5.3
Using the initial-value theorem, find the initial value of the signal corresponding to the Laplace
transform.
s +1
Y ( s) =
s( s + 2)
Verity that the answer obtained is correct.
597
Introduction to the Laplace Transform
Frequently Asked Questions linked to LO 5
ft
rrr
ut
e
t
value
rrr
Additional Solved Problems
PROBLEM 13.1
Find the Laplace transforms of the following functions.
(a) t3 1 at2 1 bt 1 3
(b) sin2 5t
(c) e5t16
t3 1 at2 1 bt 1 3
Taking Laplace transforms,
Solution (a)
3
s
s
s
s
6 2a b 3
= 4+ 3+ 2+
s
s
s
s
2
6 + 2as + bs + 3s3
=
s4
1− cos 10t 
1
1

 = L   − L  cos 10t 
L [sin 2 5t ] = L 


2

 2 
 2
1
s
= −
2 s 2( s 2 + 100)
100
=
2 s( s 2 + 100)
L[t 3 + at 2 + bt + 3] =
(b)
(c)
(d)
L [e5t +6 ] = L[e6 .e5t ] =
3!
4
+
2! a
3
e6
s −5
2
 3t
e + e−3t  



t
L [cosh 3 ] = L 
 

2

 e 6 t + e −6 t + 2 

=L


4
1 1
1
2
= 
+
+ 
4  s − 6 s + 6 s 
2
=
s 2 −18
s( s − 6)( s + 6)
+
b
2
+
(d)
cosh2 3t
598
Circuits and Networks
PROBLEM 13.2
Find the inverse transforms of the following functions.
5s + 4
(a)
(s − 1)(s + 2s + 5)
s
(c)
Solution (a)
(b)
2
(d)
2
s − 2s + 5
4s + 2
2
s + 2s + 5
s(s + 1)
2
s + 4s + 5


5s + 4

L −1 
 ( s −1)( s 2 + 2 s + 5) 
By taking partial fraction of the given function,
F ( s) =
5s + 4
( s −1)( s 2 + 2 s + 5)
=
A
BS + C
+
s −1 S 2 + 2 S + 5
Finding the values of A, B, C, we have
9
−9
11
A= ; B =
; C=
8
8
8
∴ L
−1


9


s


11 / 8 
5s + 4
9
−1 
8


=
L
−
+


 ( s −1)( s 2 + 2 s + 5) 
 8( s −1) s 2 + 2 s + 5 s 2 + 2 s + 5 
9
9
11
= e t − e−t cos 2t − e−t sin 2t
8
8
8
(b)
(c)
 4s + 4
 4s + 2 

2
 = L−1 

−
L−1  2
 s + 2 s + 5 
 ( s + 1) 2 + 22 ( s + 1) 2 + 22 


= e−t [4 cos 2t + sin 2t ]




s
s −1
1
 = L−1 

−
L−1  2
 s − 2 s + 5 
 ( s −1) 2 + 22 ( s −1) 2 + 22 


et
= [2 cos 2t + sin 2t ]
2
PROBLEM 13.3
Find the transforms of the following functions.
(a)
t e−2t sin 2t +
(c) (1 1 2t e25t)3
cos 2t
t
(b)
(d)
 s2 − 1 

 s(s + 1) 


log 
s+4
2
(s + 5s + 12)2
Introduction to the Laplace Transform
Solution (a)

cos 2t 

L t e−2t sin 2t +

t 
 cos 2t 

= L[t e−2t sin 2t ] + L 
 t 
∞
4( s + 2)
+ ∫ L(cos 2t )ds
=
[( s + 2) 2 + 4]2
s
 1 
4( s + 2)
=
+ log 

2
2
[( s + 2) + 4]
 s 2 + 4 
(b) By taking inverse transform,

  s −1
s 2 −1 

L−1  log
= L−1  log 

  s 
(
)
s
s
+
1


−1  −1 d  s −1
L
=


t 
ds  s 
1
−1  − 1 1
L
=
−

t 
s −1 s
{
(c)
} = −t1[e
+t
1
−1] = (1− et )
t
L[1 + 2t e−5t ]3 = L[1 + 8t 3e−15t + 6t e−5t + 12t 2 e−10t ]
1
83
6
2
+ 12.
= +
+
s ( s + 15) 4 ( s + 5) 2
( s + 10)3
1
48
6
24
= +
+
+
s ( s + 15) 4 ( s + 5) 2 ( s + 10)3
(d)




s+4
s+4
 = L−1 

L−1  2
2
 ( s + 5s + 12) 2 




  s + 5  + 23 

2 
4 

−1
=L
−1
=L
=e
=e


5 3


s+ +


2 2


2


2 
  s + 5  +  23  
 
2   2  


5


s+

1

 + 3 L−1 
2



2
2

2
2

 23   2
 23  


5
5


  s +  + 
 
 
  s +  + 
 

2   2  
2
2  
−5
t
2
5
 23 
− t
 t + 3 2 e 2 sin 23 t
cos 
 2 
2 23
2
−5
t
2
  23 

 
3

 cos 
sin  23  t 
  2  t +
23  2  

599
600
Circuits and Networks
PROBLEM 13.4
Find the Laplace transform of a sawtooth waveform f (t) which is periodic, with period equal to unity,
and is given by f (t) 5 at for 0 , t , 1.
Solution Laplace transform of a periodic function f (t) is
F ( s) =
1
1− e
T
−ST
∫0
f (t )e−st dt
f (t) 5 at for 0 , t , 1
Here,
∴
F ( s) =
=
=
1
1
∫ at e
1− e−ST 0
a
1 − e− s
a
1 − e− s
−st
dt
 t e−st e−st 1


 −s − s 2 

0
 e− s e− s
1 

−
+
 −s
s2
s 2 

PROBLEM 13.5
For the given function f (t) 5 3u(t) 1 2e2t, find its final value f (`) using final-value theorem.
Solution The final-value theorem is given by
lim f (t ) = lim SF ( s)
t →∞
s→
f (t) 5 3u(t) 1 2e2t
By taking Laplace transform,
∴
3
2
F ( s) = +
s s +1
2s
SF ( s) = 3 +
s +1

2s 
=3
lim SF ( s) = lim 3 +
s→ 0
s→0 
s + 
lim f (t ) = f (∞) = 3
t →∞
PROBLEM 13.6
Determine the inverse Laplace transform of the function.


4 




 2
 s + 64 




Solution
4
s + 64
2
=
(13.162)
4
s +8
2


2


Introduction to the Laplace Transform
=
601


( 4) 
8 


 2

2

8 

s +8 

 8 

1

= 
 2
2
2


s +8 

 1 8 


 4 

−1


L −1
 2
= L 
  2
2





2
s +8 



 s + 64 



(13.163)
 8 

1 −1 

L 
 2

2

2


s +8 

1
= sin 8t
2
=
(13.164)
PROBLEM 13.7
Determine the inverse Laplace transform of the function.
F (s ) =
Solution
F ( s) =
s −3
s + 4 s + 13
2
=
s −3
s
s + 4s + 13
s −3
( s + 2) + 9
2
=
( s + 2) − 5
( s + 2) 2 + 9
(13.165)
We can write the above equation as
s+2
5
−
2
( s + 2) + 9 ( s + 2) 2 + 9
By taking the inverse transform, we get



5
s + 2 
−1 

L −1 F ( s) = L −1
L
−
2
 ( s + 2) 2 + 9 

 ( s + 2) + 9 


5 −2 t
−2 t
= e cos 3t − e sin 3t
3
e−2 t
=
[3 cos 3t − 5 sin 3t ]
3
(13.166)
PROBLEM 13.8
Find the inverse transform of the following.
(a)
Solution (a)
Then
 s + 5 
 s + 6 
log 
(b)
1
2
( s + 5 2 )2
 s + 5 
Let F ( s) = log 
 s + 6 
d
1
1
d   s + 5 
−
[ F ( s)] =  log 
 =



ds
ds   s + 6  s + 5 s + 6
(13.167)
602
Circuits and Networks
We know that
d

L −1 F ( s) = −tf (t )
 ds

d

 1
1 
 = e−5t − e−6 t
∴
−
L −1  F ( s )  = L −1 
 ds

 s + 5 s + 6 
Hence,
– t f (t) 5 e–5t – e–6t
f (t ) =
Let F ( s) =
(b)
(13.168)
e −6 t − e −5 t
t
(13.169)
1
(13.170)
( s + 52 ) 2
1
1
s
= ⋅ 2
2
2 2
s ( s + 52 ) 2
(s + 5 )
2




1
s
 = L −1  1

Therefore L −1 2
2 2
 s ( s 2 + 52 ) 2 
 ( s + 5 ) 


(13.171)
According to integration theorem,
t
1

s
=
L −1
∫
2
2 2
 s ( s + 5 )  0
if
∞
Here,
 −1

s
L
 dt
2
2 

( s + 5 ) 

∞
 f (t ) 
 = ∫ F ( s ) ds
L [ f (t )] = F ( s), then L 
 t 
s
s
∫ ( s 2 + 52 ) 2
ds =
s
Therefore,
∴
or
(13.172)
∞
−1  1 
1
1

 = ⋅
2
2
2


2  s + 5 s
2 s + 52
1
1  1
f (t )
 = sin 5t
= L −1 ⋅ 2
 2 s + 52  10
0
t
t
f (t ) = sin 5t
10
t

1
s
 = t sin 5t dt
L −1 
 s ( s 2 + 52 ) 2  ∫ 10


0
(13.173)
t
=
1   − cos 5t  sin 5t 
t 

+
10   5 
25  0
=
1
[sin 5t − 5t cos 5t ]
250
PROBLEM 13.9
Find the Laplace transform of the full-wave rectified output as shown in Fig. 13.19.
(13.174)
Introduction to the Laplace Transform
603
Fig. 13.19
Solution We have
f (t ) = 10 sin vt for 0 < t <
/v
Hence,
L[ f (t )] =
∫
/ w
(e−st f (t ))dt
(13.176)
∫
(e−st 10 sin vt )dt
1− e
0
=
10
−s
1− e v
−s
v
 e−st
 /v


s
t
t
(
sin
v
v
cos
v
)
−
−
 s 2 + v2


 0

10
v e

−s 


 2

2 
v
 ( s + v )
1− e


−s 



v 
1
e
+



10v 
= 2
−s 
s + v2 
1− e v 




=
s
=
=
(13.175)
−s
1− e w
0
=

v
e 2v + e
10v
s 2 + v2
s
2
e v
−s
v

+ v

−s
2v
−s
− e 2v
 s 
cosh  
 2v 
s +v
10v
2
2
PROBLEM 13.10
Find the Laplace transform of the square wave shown in Fig. 13.20.
(13.177)
604
Circuits and Networks
Fig. 13.20
Solution We have
f (t) 5 A,
0,t,a
5 – A,
a , t , 2a
(13.178)
2a
a

 Ae−st dt + (− A)e−st dt 
∫

∫
1− e−2 as  0

a
A (1− 2e−as + e−2 as )
=
s
1− e−2 as
 as 
A
A
(1− e−as ) 2
=
= tanh  
−
as
−
as
 2 
s (1 + e )(1− e ) s
1
L [ f (t )] =
(13.179)
PROBLEM 13.11
Determine the form of the partial fraction expansion for the proper fraction.
s −1
2
(s + 9) (s + 4)(s 2 + 3s + 2)(s + 7)2
Solution
s −1
=
K1
+
( s + 9) ( s + 4)( s + 3s + 2)( s + 7)
( s + 9)
K 4 s + K5
K6
K
+ 2
+
+ 7
2
s+7
s + 3 s + 2 ( s + 7)
2
2
2
2
K
K2
+ 3
s+9 s+4
(13.180)
Alternatively, (s2 1 3s 1 2) can be factored and written as (s 1 2) (s 1 1), and the resulting partial
fraction expansion can be written as
s −1
( s + 9) ( s + 4)( s + 2)( s + 1)( s + 7)
2
2
=
K1
+
K3
K2
K
+
+ 4
( s + 9) ( s + 4) s + 2
( s + 9)
K
K6
K
+ 7
+ 5 +
2
s + 1 ( s + 7)
s+7
2
(13.181)
Introduction to the Laplace Transform
605
PROBLEM 13.12
Determine the form for a partial fraction expansion for the improper fraction.
3
2
6s + 100s + 85s + 52
3
2
s + 7s + 14s + 8
Solution Because the expression is not a proper fraction, it cannot be expanded into partial fractions.
However, if the denominator is divided into the numerator, a part of the expression becomes a proper fraction,
and that part can be expanded.
6 s3 + 100 s 2 + 85s + 52
s3 + 7s 2 + 14 s + 8
∴
6 s3 + 100 s2 + 85s + 52
s + 7 s + 14 s + 8
3
2
= 6+
58s 2 + s + 4
s3 + 7 s 2 + 14 s + 8
= 6+
58s 2 + s + 4
( s + 1)( s + 2)( s + 4)
(13.182)
= 6+
K
K1
K
+ 2 + 3
s +1 s + 2 s + 4
(13.183)
PROBLEM 13.13
Determine the inverse Laplace transform of F(s), where
F (s ) =
s
(s + 1)2 (s + 4)
.
Solution From the rules given for expanding proper fractions,
s
( s + 1) ( s + 4)
2
=
K1
( s + 1)
2
+
K
K2
+ 3
s +1 s + 4
(13.184)
K1 = F ( s) ( s + 1)2 s=1
s
−1
=
s=−1 =
3
s+4
K3 = F ( s) ( s + 4) s=−4
s
−4
=
=
2
9
( s + 1)
We can determine K2 by putting the right side of the equation on a common denominator and equating
numerators.
K1 ( s + 4) + K 2 ( s + 1)( s + 4) + K3 ( s + 1) 2
s
(13.185)
=
( s + 1) 2 ( s + 4)
( s + 1) 2 ( s + 4)
−1
−4
and K 3 =
3
9
Equating numerators results in
Since
K1 =
 −11
 
 −16

4
+ 4 K 2 
s = 
+ 5 K 2  s +  K 2 −  s 2 + 
 9


 
 9

9
606
Circuits and Networks
Equating coefficients results in
−11
+ 5K 2 = 1
9
4
K2 =
9
s
F ( s) =
L
=
( s + 1) 2 ( s + 4)
−1
{ F ( s)} = L
 −1
 
 3 
( s + 1) 2
+
( 4 / 9) (−4 / 9)
+
s +1
s+4

 −1



 3
( 4 / 9) (−4 / 9) 
+
+



s +1
s + 4 
( s + 1) 2







−1
From the rules,
f (t ) =
−1 −t 4 −t 4 −4t
te + e − e
3
9
9
(13.186)
PROBLEM 13.14
Expand the following proper fraction into partial fraction.
F (s ) =
1
2
(s + 3s + 2)(s + 4)
Solution By expanding proper fractions,
1
=
K1s + K 2
K3
s+4
2
+
K 3 = F ( s) ( s + 4)
s=−4
( s + 3s + 2)( s + 4)
2
=
( s + 3s + 2)
(13.187)
1
( s 2 + 3s + 2)
s=−4
1
6
Putting the right-hand side of the expanded equation over a common denominator to determine K1 and K2
results in
1
K1s + K 2
1
6
=
+
( s 2 + 3s + 2) ( s + 4) ( s 2 + 3s + 2) s + 4
1
1
K1s 2 + 4 s K1 + K 2 s + 4 K 2 +   s 2 +   s + 1 / 3
 6 
 2 
=
2
( s + 3s + 2) ( s + 4)




 K1 + 1  s 2 + 4 K1 + 4 K 2 + 1  s + ( 4 K 2 + 1 / 3)


2 
6 
(13.188)
=
( s 2 + 3s + 2) ( s + 4)
K3 =
Introduction to the Laplace Transform
607
Equating numerators, we get
1
K1 + = 0
6
1
4 K1 + K 2 + = 0
2
1
4K2 + = 1
3
From the above equations, we get
−1
1
K1 =
, K2 =
6
6
The required partial fraction expansion is
   
 −1 s +  1 
 6   6 
 1 
 
 6 
F ( s) = 2
+
s + 3s + 2 s + 4
(13.189)
PROBLEM 13.15
Determine the inverse Laplace transform of the following function.
F (s ) =
96(s 2 + 17s + 60)
s 3 + 14s 2 + 48 s
Solution The function F(s) can be factorised as given.
F ( s) =
96( s + 5)( s + 12)
s( s + 8)( s + 6)
(13.190)
K
K1
K
+ 2 + 3
s
s +8 s + 6
To find K1, we multiply both sides by s and then put s 5 0
96 ( s + 5) ( s + 12)
K1 = F ( s) s s=0 =
s=0 = 120
( s + 8) ( s + 6)
F ( s) =
To find the value of K2, we multiply both sides by s 1 8 and then evaluate both sides at s 5 2 8
96( s + 5) ( s + 12)
K 2 = F ( s) ( s + 8) s=−8 =
s=−8 = −72
s ( s + 6)
Then K3 is
96 ( s + 5) ( s + 12)
s( s + 8)
s=−6
= K3 = 48
Therefore,
96( s + 5) ( s + 12) 120
48
72
=
+
−
s( s + 8) ( s + 6)
s
s + 6 s +8
(13.191)
608
Circuits and Networks
By taking inverse transform of the above function, we get

120
 96( s + 5) ( s + 12) 
48
72 
−1

L −1
+
−



= L 



+
s
(
s
+
8
)
(
s
+
6
)
s
s
6
s
+ 8 






5 120 1 48e–6t – 72e–8t
(13.192)
PROBLEM 13.16
Determine the inverse Laplace transform of the following function.
F (s ) =
100 (s + 3)
(s + 6)(s 2 + 6s + 25)
Solution By factoring denominator, we have
100 ( s + 3)
( s + 6) ( s + 6 s + 25)
2
=
K3
K1
K2
+
+
s + 6 s + 3− j4 s + 3 + j4
(13.193)
To find K1, K2, and K3, we use the same process as before:
K1 =
K2 =
K3 =
100 ( s + 3)
s + 6 s + 25
2
s=−6
=
100 ( s + 3)
( s + 6) ( s + 3 + j 4)
2
100 ( s + 3)
( s + 6) ( s + 3 − j 4)
2
100 (−3)
= −12
25
s=−3+ j 4
=
100 ( j 4)
(3 + j 4) ( j8)
s=−3− j 4
=
100 (− j 4)
(3 − j 4)(− j8)
Then
100 ( s + 3)
( s + 6) ( s + 6 s + 25)
2
=
−12 10 ∠− 53.13° 10 ∠53.13°
+
+
s+6
s + 3− j4
s + 3 + j4
(13.194)
By taking inverse Laplace transform, we get



100 ( s + 3)
−6 t
− j 53.13°

L −1
⋅ e−( 3− j 4 )t

 = −12e + 10e
2


(
s
+
6
)
(
s
+
6
s
+
25
)




1 10e j53.13º ? e–(3 1j4)t
By simplifying, we get




100 ( s + 3)

L −1
= −12e−6t + 20e−3t cos ( 4t − 53.13°)


2



 ( s + 6) ( s + 6 s + 25) 

{
}
(13.195)
609
Introduction to the Laplace Transform
PROBLEM 13.17
Obtain inverse Laplace transform of the following function.
F (s ) =
100 (s + 25)
s (s + 5)3
Solution By factorising the denominator, we have
100( s + 25)
s( s + 5)
3
=
K3
K1
K2
K
+
+
+ 4
3
2
+5
s
s
( s + 5)
( s + 5)
(13.196)
We find K1, as
K1 =
100 ( s + 25)
( s + 5)3
s= 0
= 20
To find K2, we multiply both sides by (s 1 5)3 and then evaluate both sides at –5.
100 ( s + 25)
s
∴
s=−5
=
K1 ( s + 5)3
s
s=−5
+ K 2 + K 3 ( s + 5)
s=−5
+ K 4 ( s + 5) 2
s=−5
K2 5 –400
To find K3, we first multiply both sides by (s 1 5)3. Next we differentiate both sides once with respect to
s and then evaluate at s 5 – 5.
d 100 ( s + 25) 
d  K ( s + 5)3 
d


+ [ K 2 ]s =−5
=  1

 s =−5 ds 
ds 
s
s
ds
 s =−5

d
d
+ [ K 3 ( s + 5)]s =−5 + [ K 4 ( s + 5) 2 ]s =−5
ds
ds
 s − ( s + 25) 

100 
= K 3 = −100

 s =−5
s2
To find K4, we first multiply both sides by (s 1 5)3. Next we differentiate both sides twice with respect to s
and then evaluate both sides at s 5 –5. After simplifying the first derivative, the second derivative becomes
100
d  −25 
d  ( s + 5) 2 ( 2 s − 5) 
d
 2 
+ 0 + [ K 3 ]s=−5
= K1 

ds  s  s=−5
ds 
ds
s2
 s=−5
d
+ [2 K 4 ( s + 5)]s=−5
ds
or
– 40 5 2K4
K4 5 – 20
Then
100 ( s + 25)
s ( s + 5)
3
=
20
400
100
20
−
−
−
3
2
s ( s + 5)
s+5
( s + 5)
(13.197)
610
Circuits and Networks
By taking inverse transform, we get
100 ( s + 25) 

 = 20 − 200t 2 e−5t −100te−5t − 20e−5t
L −1

3 


s
(
s
+
5
)




(13.198)
PROBLEM 13.18
Verify the initial- and final-value theorems for the function.
f (t) 5 e–t (sin 3t 1 cos 5t)
–t
Solution f (t) 5 e (sin 3t 1 cos 5t)
F(s) 5 L [ f (t)] 5 L [e–t (sin 3t 1 cos 5t)]
Since L(e−t sin 3t ) =
and
L (e−t cos 5t ) =
∴
(13.199)
(13.200)
3
( s + 1) 2 + 32
s +1
( s + 1) 2 + 52
F ( s) = L [ f (t )] =
3
( s + 1) + 3
2
2
+
s +1
( s + 1) 2 + 52
(13.201)
According to initial-value theorem,
Lt f (t ) = Lt SF ( s)
t →0
s→∞
F ( s) =
3
+
s +1
s + 2 s + 10 s 2s + 26
s2 + s
3s
+
SF ( s) =
 2 10 
 2 26 
s 2 1 + + 2  s 2 1 + + 2 


s s 
s s 
1
1
3
+
+
=
 2 26 
 2 10 
2
26
s 1 + + 2  1 + + 2 s 1 + + 2 


8 s
s s 
s s 
lim SF ( s) = 1
2
2
s→∞
f (t) 5 e –t (sin 3t 1 cos 5t)
lim f (t ) = 1
t →0
Hence, the initial-value theorem is proved.
According to the final-value theorem,
lim f (t ) = lim SF ( s)
t →∞
s→ 0
lim SF ( s) = 0
s→ 0
lim f (t ) = 0
t →∞
Hence, the final-value theorem is proved.
(13.202)
Introduction to the Laplace Transform
611
PROBLEM 13.19
Find the value of i(01) using the initial-value theorem for the function given
I (s ) =
2s + 3
(s + 1)(s + 3)
Verify the result by solving it for i(t).
Solution The initial-value theorem is given by
lim i(t ) = lim SI ( s)
t →0
s→∞
= lim
s( 2 s + 3)
(13.203)
s→∞ ( s + 1) ( s + 3)
Taking S common and putting S 5 `, we get

3
s 2 2 + 

s
=2
lim



s→∞ 2
1
3
s 1 + 1 + 
 s  s 
(13.204)
To verify the result, we solve for i(t) and put t → `.
Taking partial fractions,
I ( s) =
where
∴
A
B
+
s +1 s + 3
A = ( s + 1)
2s + 3
1
=
( s + 1) ( s + 3) s=−1 2
B = ( s + 3)
3
2s + 3
=
( s + 1) ( s + 3) s=−3 2
I ( s) =
1
3
+
2( s + 1) 2( s + 3)
(13.205)
Taking inverse transform, we get
1
3
i(t ) = e−t + e−3t
2
2
(13.206)
Answers to Practice Problems
13-2.1
f (t) 5 sin
13-2.3
(a)
13-2.5
t
[u(t) – u(t – 4)]
2
I 5 26
−e
−s
(b)
−s
(−0.5e )[log( 0.5e−s )]2
I50
(c) I 5 0
612
13-2.8
Circuits and Networks
(a)
2
( s + a)
3
(b)
Sb
( s + a) − b
2
2
(c)
S 2 − v2
( s 2 + v2 ) 2
d
d(t ) + 2d(t ) − 2e−3t u(t ) + 4e−2t u(t )
dt


25
13-2.12 f (t) 5 e25t 10 −
sin 12t 


3
13-2.11
y (t ) =
13-2.14
(a)
(b)
f (t) 5 5 d(t) 2 3e22t 1 2e24t
(c)
f (t) 5 2
13-5.2
d
d(t) 2 2d(t) 1 4e24t
dt
1
X ( s) = [1− 3e−ST + 4e−2 ST − 4e−4 ST + 2e−5ST ]
s
13-3.1
13-4.3
f (t) 5 220e22t (t cos t 2 sint)
f (t) 5 t e22t u(t)
(a)
f (0) 5 7, f (`) 5 0
(b)
f (0) 5 4, f (`) 5 1
(c)
f (0) 5 0, f (`) 5 0
13-5.3
Lt SY ( s) = 1
s→∞
Objective-Type Questions
rrr 13.1
Laplace transform analysis gives
(a) time-domain response only
(c) both (a) and (b)
rrr 13.2
The Laplace transform of a unit step function is
(a)
rrr 13.3
1
s
(b) 1
s
F ( s)
s2
(d)
s+a
(b) SF(s) – f (0)
(c) F(s) – f (0)
(d) f (0)
(b) SF(s) – f (0)
(c) F (s) – f (0)
(d) f 9(0)
 s
(c) F  

(d) F(s – 5)
(c) u(t – a)
(d) zero
The Laplace transform of e5t f (t) is
(a) F(s)
rrr 13.6
1
The Laplace transform of the integral of function f (t) is
(a)
rrr 13.5
(c)
The Laplace transform of the first derivative of a function f (t) is
(a) F(s)/s
rrr 13.4
(b) frequency-domain response only
(d) none
(b) F(s – 1)
The inverse Laplace transform of
(a) u(t) – u(t – a)
 
1
(1 – e–as) is
s
(b) u(t)
613
Introduction to the Laplace Transform
rrr 13.7
6
The inverse transform of
S4
is
(a) 3
(b) t2
s+
rrr 13.8 The inverse transform of 2 log 
 s
(a)
rrr 13.9
2 − e−2 t
t
(b)
(c) t3

 is

e−2t
t
(c)
(d) 3t
2
t
(d)
2 + e−2 t
t
The Laplace transform of a square wave with amplitude of peak value A and period T is
(a)
1 + e−sT
(b)
1− e−sT
A 1− e−sT 


s 1 + e−sT 
rrr 13.10 The inverse Laplace transform of the function
(a) 2et – e–3t
(b) 2e–t 1 e–3t
(c)
A 1 + e sT 


s 1− e sT 
(d)
A 1− e sT 


s 1 + e sT 
s+5
is
( s + 1) ( s + 3)
(c) e–t – 2e–3t
(d) e–t 1 2e–3t
rrr 13.11 The Laplace transform of a unit ramp function at t 5 a is
(a)
1
( s + a)
( s + a)
2s + 1
2
rrr 13.12 The initial value of
(a) 20
e25t
2( s + 1)
s + 2s + 5
2
(d)
a
s2
is
(c) zero
(d) 1
(c) 10
(d) 25
is
(b) 19
2
e−as
s
(b) infinite
rrr 13.13 The initial value of 20 – 10t –
L [ f (t )] =
(c)
2
s 4 + 8s3 + 16 s 2 + s
(a) 2
rrr 13.14
e−as
(b)
, then f (0 1) and f (`) are given by
(a) 0, 2 respectively
(b) 2, 0 respectively
(c) 0, 1 respectively
(d)
2
, 0 respectively
5
rrr 13.15 The final-value theorem is used to find the
(a) steady-state value of the system output
(c) transient behaviour of the system output
(b) initial value of the system output
(d) none of these
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14
LEARNING OBJECTIVES
The Laplace transform is an attractive tool in circuit analysis. It transforms a set of linear constantcoefficient differential equations into a set of linear polynomial equations. It automatically introduces into
the polynomial equations the initial values of the current and voltage variables. In the circuit analysis, we can
develop the s-domain circuit models for various elements and s-domain equations can be written directly.
14.1
CIRCUIT ELEMENTS IN THE S-DOMAIN
For any element, we write the time-domain equation that relates the terminal
LO 1
voltage to the terminal current. Then, we take the Laplace transform of the timedomain equation. This gives an algebraic relation between the s-domain current
and voltage. The dimensions of a transformed voltage is volt- seconds, and the
dimension of a transformed current is ampere-seconds. A voltage-to-current ratio in the s-domain carries
the dimension of volts per ampere. An impedance in the s-domain is measured in ohms, and admittance is
measured in Siemens.
14.1.1 A Resistor in the s-Domain
Consider the resistive element shown in Fig. 14.1 From Ohm’s law,
v 5 Ri
The Laplace transform of Eq. (14.1) is
V 5 RI
where
V 5 L [v] and I 5 L [i]
(14.1)
(14.2)
615
Application of the Laplace Transform in Circuit Analysis
Equation (14.2) states that the s-domain
equivalent circuit of a resistor is simply a resistance
of R ohms that carries a current of I ampere seconds
and has a terminal voltage of V volt-seconds.
Figures 14.1 (a) and (b) show the time- and
Fig. 14.1
frequency-domain circuits of the resistor respectively.
The resistance element does not change while going from the time domain to the frequency domain.
14.1.2 An Inductor in the s-Domain
Consider an inductor shown in Fig. 14.2 with an initial current of I0 amperes.
The time domain relation between voltage and current is
Fig. 14.2
di
v=L
dt
(14.3)
The Laplace transform of Eq. (14.3) gives
V 5 L [SI – i(0)] 5 SLI – LI0
The above equation satisfies two circuits. The
first consists of an impedance of SL ohms in
series with an independent voltage source of LI0
volt-seconds as shown in Fig. 14.3 (a).
The second s-domain equivalent circuit that
satisfies Eq. (14.4) consists of an impedance of
SL ohms in parallel with an independent current
source of I0/S ampere- seconds, as shown in
Fig. 14.3 (b).
By solving Eq. (14.4) for the current I, we can
construct the circuit shown in Fig. 14.3 (b).
I=
V + LI 0
I
V
=
+ 0
SL
SL S
Fig. 14.4
(14.4)
(a)
Fig. 14.3
(b)
(14.5)
If the initial energy stored in the inductor is zero, i.e., if I0 5 0, the
s-domain equivalent circuit of the inductor reduces to an inductor with an
impedance of SL ohms as shown in Fig. 14.4.
14.1.3 A Capacitor in the s-domain
Consider an initially charged capacitor shown in Fig. 14.5. The initial
voltage on the capacitor is V0 volts.
The voltage–current relation in the time domain is
dv
Fig. 14.5
i=c
(14.6)
dt
By taking Laplace transforms both sides, we get
I 5 C [SV – v(0)]
I 5 SCV – CV0
(14.7)
616
Circuits and Networks
The Equation 14.7 represents two circuits. First, the parallel equivalent
circuit for capacitor initially charged to V0 volts is shown in Fig. 14.6.
Secondly, the series equivalent circuit can be derived for the charged
capacitor by solving Eq. (14.7) for V.
 1 
V
V =   I + 0
 SC 
S
(14.8)
Figure 14.7(a) shows the circuit that satisfies Eq. (14.8).
The s-domain circuit for a capacitor when the initial voltage is zero is
shown in Fig. 14.7 (b).
Fig. 14.6
Fig. 14.7 (a)
Fig. 14.7 (b)
Frequently Asked Questions linked to LO 1*
rrr
s
14.2
APPLICATIONS
In this section, we illustrate how to use the Laplace transform to determine
the transient behaviour of several linear lumped-parameter circuits. In analysis
of familiar circuits, the Laplace transform approach yields the same results
like the time-domain analysis. In all the examples, the ease of manipulating
algebraic equations instead of differential equations should be apparent.
LO 2
14.2.1 Natural Response of an RC Circuit
In this section, we find the natural response of an RC circuit through Laplace transform techniques. Consider
the capacitor discharge circuit shown in Fig. 14.8. Assume the capacitor is initially charged to V0 volts. The
series equivalent s-domain circuit is shown in Fig. 14.9.
Fig. 14.8
Fig. 14.9
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
Application of the Laplace Transform in Circuit Analysis
617
From the circuit shown in Fig. 14.9(b), applying Kirchhoff’s voltage law around the loop, we have
V0
1
(14.9)
=
I + RI
S
SC
Solving for the above equation yields
CV0
V /R
= 0
 1 
RCS + 1

S+
 RC 
By taking the inverse transform of Eq. (14.10), we get
−t
V
i = 0 e RC
R
We can determine v by simply applying Ohm’s law from the circuit
I=
(14.10)
(14.11)
−t
v = Ri = V0 e RC
(14.12)
Now we can use the parallel equivalent circuit of Fig. 14.9 (a).
Figure 14.10 shows the new s-domain equivalent circuit.
By taking node voltage equation, we get
V
+ SCV = CV0
R
Fig. 14.10
(14.13)
Solving Eq. (14.13) for V gives
V=
V0
(14.14)
1
S+
RC
By taking inverse transform, we get
−t
v = V0 e RC = V0 e
−t
t
(14.15)
where t is the time constant t 5 RC
14.2.2 Step Response of a Parallel Circuit
Consider the parallel RLC circuit shown in Fig. 14.11. We can find the expression for iL after the constant
current source is switched across the parallel elements.
The initial energy stored in the circuit is zero.
The s-domain equivalent circuit is shown in Fig. 14.12.
Here, an independent source can be transformed easily
from the time domain to the frequency domain. Opening
the switch results in a step change in the current applied
Fig. 14.11
to the circuit.
By applying Kirchhoff’s current law, we get
SCR +
I
V V
+
= dc .
R SL
S
(14.16)
618
Circuits and Networks
Solving Eq. (14.16) for V gives
I dc / C

1 
1
S 2 + 
S+
 RC 
LC
We know the current in inductor IL
Fig. 14.12
V
IL =
SL
Substituting Eq. (14.17) into Eq. (14.18) gives
V=
I dc / LC
 2  1 

S + 1 
S  S + 
 RC 


LC 
IL =
(14.17)
(14.18)
(14.19)
By taking the inverse transform, we can obtain IL.
14.2.3 Transient Response of a Parallel RLC Circuit
The transient behaviour of a circuit arises from replacing the dc current source in the circuit shown in Fig.
14.11 with a sinusoidal current source. The new current source is
ig 5 Im cos vt
(14.20)
The s-domain expression for the source current is
SI M
Ig =
S + v2
2
(14.21)
The voltage across the parallel elements is
V=
( I g / C )S
 1 
 1 
S 2 + 
S + 
 RC 
 LC 
(14.22)
Substituting Eq. (14.21) into Eq. (14.22) results in
V=
( I m / C )S 2

 1 
 1 

( S 2 + v2 )  S 2 + 
S + 
 RC 
 LC 


(14.23)
from which
IL =
V
=
SL
( I m / LC )S

 1 
 1 

( S 2 + v2 )  S 2 + 
S + 
 RC 
 LC 


(14.24)
14.2.4 Use of Thevenin’s Equivalent
In this section, we show how to use Thevenin’s equivalent in the s-domain. Consider a circuit shown in Fig.
14.13. We find the capacitor current that results from closing the switch. The energy stored in the circuit prior
to closing is zero.
619
Application of the Laplace Transform in Circuit Analysis
Fig. 14.13
Fig. 14.14
To find ic, we first construct the s-domain equivalent circuit and the find the Thevenin equivalent of this
circuit with respect to the terminals of the capacitor. Figure 14.14 shows the s-domain circuit.
The open-circuit voltage across terminals a, b is
VTh
 480 

(0.0025)
 S 
480
=
=
20 + 0.002S
S +104
(14.25)
The Thevenin impedance seen from terminals a and b equals the 60 V resistor in series with the parallel
combination of the 20 V resistor and the 2 mH inductor.
Thus,
ZTh = 60 +
0.002S ( 20) 80( S + 7500)
=
20 + 0.002S
S + 104
(14.26)
A simplified version of the Thevenin equivalent circuit is shown in Fig. 14.15.
Fig. 14.15
Thus, the capacitor current IC equals the Thevenin voltage divided by the total series impedance.
Thus,
IC =
480 / ( S + 104 )
80( S + 7500) / ( S + 104 ) + ( 2×105 ) / S  


 
(14.27)
We simplify Eq. (14.27) to
IC =
=
6S
S + 10, 000 S + 25×106
6S
2
( S + 5000) 2
(14.28)
620
Circuits and Networks
By taking partial fraction expansion, we get
IC =
−3000
( S + 5000)
2
+
6
( S + 5000)
(14.29)
By taking inverse transform, we get
ic 5 (– 30,000t e–5000t 1 6 e–5000t) A
(14.30)
Now the voltage across capacitor is
VC =
IC
2×105
6S
12×105
=
=
SC
S ( S + 5000) 2 ( S + 5000) 2
(14.31)
By taking inverse transform, we get
vc 5 12 3 105 t e25000t
(14.32)
14.2.5 Circuit with Mutual Inductance
In this section, we illustrate an example how to use the Laplace transform to analyze the transient response of
a circuit that contains mutual inductance as shown in Fig. 14.16.
Fig. 14.16
To make-before break switch has been in position ‘a’ for a long time. At t 5 0, the switch moves
instantaneously to the position b. The problem is to derive the time- domain expression for i2.
We begin by redrawing the circuit in Fig. 14.16, with the switch in the position b and the magnetically
coupled coils replaced with a T equivalent circuit as shown in Fig. 14.17.
The s-domain equivalent circuit for the circuit of Fig. 14.17 is shown in Fig. 14.18.
The initial currents are
60
= 5A
12
i2(0) 5 0
i1 (0) =
(14.33)
(14.34)
The initial value of the current in the 2 H inductor is
i1(0) 1 i2(0) 5 5 A
(14.35)
The s-domain mesh equations in Fig. 14.18 are
(3 1 2S ) I1 1 2SI2 5 0
(14.36)
2SI1 1 (12 1 8S )I2 5 10
(14.37)
Application of the Laplace Transform in Circuit Analysis
621
Fig. 14.17
Fig. 14.18
Solving for I2 yields
2.5
I2 =
( S + 1) ( S + 3)
(14.38)
By taking partial fraction expansion gives
1.25 1.25
I2 =
−
S +1 S + 3
By taking inverse transform of Eq. (14.39) gives
i2 5 (1.25 e2t – 1.25 e23t) A
(14.39)
(14.40)
14.2.6 Use of Superposition
Consider a circuit shown in Fig. 14.19 having two sources and the inductor is carrying and initial current iL (0)
amperes and the capacitor is carrying an initial voltage of vc (0) volts. The desired response of the circuit is the
voltage across the resistor R2, labeled v2.
The s-domain equivalent circuit for Fig. 14.19 is shown in Fig. 14.20. Here, we have taken parallel
equivalents for L and C into consideration. Now we find V2 using node-voltage method.
To find V2 by superposition, we calculate the voltage V2 resulting from each source acting alone, and then
we sum the voltages. We consider Vg acting alone by setting the other three current sources equal to zero.
Figure 14.21 shows the resulting circuit.
Fig. 14.19
622
Circuits and Networks
Fig. 14.20
V19 and V29 are the voltages across the inductor and
resistor when Vg acting alone.
The two equations described in the circuit in Fig.
14.21 are
1

 + 1 + SC V ′− SCV ′ = Vg
 1
2
 R
SL
R1

1
(14.41)
Fig. 14.21

1
−SC V1′+  + SC V2′ = 0

 R2
The above equations can be written as
Vg
Y11 V1′+ Y12 V2′ =
R1
Y21 V19 1 Y22 V29 5 0
(14.43)
(14.44)
1
1
+ + SC
R1 SL
Y12 5 – SC 5 Y21
1
Y22 =
+ SC
R2
where Y11 =
Solving Eqs (14.43) and (14.44) for V29 gives
V2′ =
−Y12 /R1
Y11 Y22 − Y122
Vg
(14.45)
With the current source Ig acting alone, the circuit
shown in Fig. 14.20 reduces to the one shown in
Fig. 14.22.
The two node-voltage equations are given by
Y11 V10 1 Y12 V20 5 0
(14.42)
(14.46)
Fig. 14.22
623
Application of the Laplace Transform in Circuit Analysis
Y21 V10 1 Y22 V20 5 Ig
(14.47)
Solving for V20 yields
Y11
V2′′=
Ig
Y11 Y22 − Y122
(14.48)
The circuit shown in Fig. 14.23 gives when the
energised inductor acting alone on the circuit of Fig.
14.20.
The two node-voltage equations are given by
i ( 0)
(14.49)
Y11 V1′′′ + Y12 V2′′′ = − L
S
(14.50)
Y V ′′′ + Y V ′′′ = 0
21 1
22
2
Thus,
Fig. 14.23
V2′′′ =
Y12 /S
Y11 Y22 − Y122
− iL (0)
(14.51)
The circuit shown in Fig. 14.24 gives when the energy stored in the capacitor acting alone.
The node-voltage equations describing this circuit are
Y11 V100 1 Y12 V200 5 vc(0)C
(14.52)
Y21 V100 1 Y22 V200 5 – vc(0)C
(14.53)
Solving for V200 yields
V 00 =
−(Y11 + Y12 )C
Y11 Y22 − Y122
vc (0)
(14.54)
The expression for V2 is
V2 5 V291V20 1 V21V200
=
Fig. 14.24
−(Y12 /R1 )
Y11 Y22 − Y122
+
Y12 /S
Vg +
Y11
Y11 Y22 − Y122
iL (0) +
Y11 Y22 − Y122
Ig
−C (Y11 + Y12 )
vc (0)
Y11 Y22 − Y122
By taking inverse transform, we can obtain time domain voltage across the resistor R2.
(14.55)
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 2*
A 500 V resistor, a 16 mH inductor, and a 25 nF capacitor are connected in parallel which is
placed in series with a 2000 V resistor. Express the impedance of this series combination as a
rational function of s.
rrr 14-2.2 A 1 kV resistor is in series with a 500 mH inductor. This series combination is in parallel with
a 0.4 mF capacitor. Express the equivalent s-domain impedance of these parallel branches as a
rrr 14-2.1
*Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
624
rrr 14-2.3
Circuits and Networks
rational function.
The energy stored in the circuit shown is zero at the time when the switch is closed.
(a) Find the s-domain expression for I.
(b) Find the s-domain expression for i when t > 0.
(c) Find the s-domain expression for V.
(d) Find the time-domain expression for v when t > 0.
Fig. Q.3
rrr 14-2.4
The dc current and voltage sources are applied simultaneously to the circuit shown. No energy
is stored in the circuit at the instant of application.
(a) Derive the s-domain expressions for V1 and V2.
(b) For t > 0, derive the time-domain expressions for v1 and v2.
(c) Calculate v1(01) and v2(01).
(d) Compute the steady state value of v1 and v2.
rrr 14-2.5
The energy stored in the circuit shown is zero at the instant the two sources are turned on.
(a) Find the component of v for t > 0 owing to the voltage source.
(b) Find the component of v for t > 0 owing to the current source.
(c) Find the expression for v when t > 0.
rrr 14-2.6
In the circuit shown in Fig. Q.6, there is no energy stored at the time the current source turns on.
Given that ig 5 100 u(t) A;
Fig. Q.4
Fig. Q.5
625
Application of the Laplace Transform in Circuit Analysis
(a) Find I0(s).
(b) Use the initial- and final value theorems to find i0(01) and i0(`).
(c) Determine if the results obtained in (b) agree with known circuit behaviour.
(d) Find i0(t).
rrr 14-2.7 There is no energy stored in the circuit seen in Fig. Q.7 at the time the two sources are
energised.
(a) Use the principle of superposition to find V0.
(b) Find v0 for t > 0.
Fig. Q.6
rrr 14-2.8
rrr 14-2.9
Fig. Q.7
Find (a) the unit step, and (b) the unit impulse response of the circuit shown in Fig. Q.8.
There is no energy stored in the circuit shown in Fig. Q.9 at the time the impulse voltage is
applied. Find v0(t) for t $ 0.
Fig. Q.8
Fig. Q.9
The switch in the circuit shown in Fig. Q.10 has been in position a for a long time. At t 5 0, the
switch moves to the position b. Compute (a) v1(0) (b) v1(0 –) (c) v3(0 –) (d) i(t) (e) v1 (01) (f) v2
(01) (g) v3 (01).
rrr 14-2.11 Find v0 (t) in the circuit shown in Fig. Q.11, is 5 5 u(t) A, Using PSpice.
rrr14-2.10
Fig. Q.10
Fig. Q.11
the RLC circuit shown in Fig. Q.12, find the complete response if v (0) 5 2 V when the
switch is closed. Use PSpice.
rrr 14-2.12 For
626
Circuits and Networks
Fig. Q.12
Frequently Asked Questions linked to LO 2
a
rr
1W a
b
b
t
it
1W
t=0
10 V
3 F Vc(t)
i(t)
6F
Fig. Q.1
a
rr
t
b
1
F
2
a
1W
2W
b
1W
10 V
1
F
2
–1
e u(t)
2H
Fig. Q.2
V2 (s)
.
V1 (s)
rr
1W
1W
Io
V1
2V1
1W
2Io
V2
Fig. Q.3
14.3
TRANSFER FUNCTION
The transfer function is defined as the s-domain ratio of the Laplace transform of
the output (response) to the Laplace transform of the input (source). In computing
the transfer function, we restrict our attention to circuits where all initial conditions
are zero. If a circuit has a multiple independent sources, we can find the transfer
LO 3
V t
Application of the Laplace Transform in Circuit Analysis
627
function for each source and use superposition to find the response to all sources.
The transfer function is
Y (S )
(14.56)
H (S ) =
X (S )
where Y(S ) is the Laplace transform of the output signal, and
X(S ) is the Laplace transform of the input signal. Note that the
transfer function depends on what is defined as the output signal.
Consider a series circuit shown in Fig. 14.25.
If the current is defined as the response signal of the circuit
then the transfer function
I
1
SC
(14.57)
H (S ) =
=
= 2
1
V
S
LC
+
RCS
+
1
g
Fig. 14.25
R + SL +
SC
In the above equation, we recognised that I corresponds to the output Y(S ) and Vg corresponds to the input
X(S ). If the voltage across the capacitor is defined as the output signal of the circuit in Fig. 14.25, the transfer
function is
H (S ) =
V
=
Vg
1/SC
R + SL +
1
SC
=
1
S LC + RCS + 1
2
(14.58)
Thus, because circuits may have multiple sources and because the definition of the output signal of interest
can vary, a single circuit can generate many transfer functions. When multiple sources are involved, no single
transfer function represent the total output-transfer functions associated with each source must be combined
using superposition to yield the total response. We can write the circuit output as the product of the transfer
function and the driving function
Y(S ) 5 H(S ) X(S )
(14.59)
H(S ) is a rational function of S and X(S ) is also a rational function of S for the excitation functions of most
interest in circuit analysis. We can expand the right-hand side of Eq. (14.58) into a sum of partial fractions.
14.4
USE OF TRANSFER FUNCTION IN CIRCUIT ANALYSIS
LO 3
Consider the response of the circuit to a delayed input. If the input is delayed by a seconds
L [x(t – a) u(t – a)] 5 e2as2X(S )
(14.60)
The response becomes
Y(S ) 5 H(S ) X(S )e2as
(14.61)
If y(t) 5 L21 [H(S ) X(S )], then from Eq. 14.61,
y(t – a) u(t – a) 5 L21 [H(S ) X(S ) e2as]
(14.62)
Therefore, delaying the input by a seconds simply delays the response function by a seconds. A circuit that
exhibits this characteristic is said to be time invariant.
If a unit impulse source drives the circuit, the response of the circuit equals the inverse transform of the
628
Circuits and Networks
transfer function. Thus, if
x(t) 5 d(t), then X(S ) 5 1
and
Y(S ) 5 H(S )
(14.63)
Hence, y(t) 5 h(t)
(14.64)
where the inverse transform of the transfer function equals the unit impulse response of the circuit. The unit
impulse response of the circuit h(t) contains enough information to compute the response to any source that
drives the circuit.
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 3
Derive the numerical expression for the transfer
function v0/Ig for the circuit shown in Fig. Q.1.
r 14-3.2 The unit impulse response of a circuit is
v0(t) 5 10,000 e–70t cos (240 t 1 u) u(t) V
7
where tan u =
24
(a) Find the transfer function of the circuit.
(b) Find the unit step response of the circuit.
r 14-3.1
Fig. Q.1
Frequently Asked Questions linked to LO 3
C1
R1
rr
V2 (s)
1
=
V1 (s) R1C2
R2
V1
S
+
R
C
R
1
2 (C1 + C2 )
+
S 2 + R1C2 + 1 2
R1 R2C1C2
R1 R2C1C2
C2
V2
Fig. Q.1
r
V2 (s) b3 s3 + b2 s2 + b1s + b0
=
V1 (s) a3 s3 + a2 s2 + a1s + a0
a b R R R C C C
a R R C C C
R R C C
b R R R C C
a R C C
RC
C C R
b R C C
a b
14.5
R1
V1
R R C C
R2
C1
C2
C3
R3
Fig. Q.2
THE TRANSFER FUNCTION AND THE CONVOLUTION INTEGRAL
The convolution integral relates the output y(t) of a linear time invariant circuit
to the input x(t) of the circuit and the circuits impulse response h(t). The
convolution integral is defined as
∞
y (t ) =
∫
−∞
∞
h( t) x(t − t) d t =
∫
−∞
h(t − t) x( t) d t
(14.65)
LO 4
V2
Application of the Laplace Transform in Circuit Analysis
629
The above equation is based on the assumption that the circuit is
linear and time invariant. Because the circuit is linear, the principle of
superposition is valid, and because it is time invariant, the amount of the
response delay is exactly the same as that of the input delay. Consider
Fig. 14.26
block diagram of a general circuit shown in Fig. 14.26 in which h(t)
represents any linear time-invariant circuit whose impulse response is known, x(t) represents the excitation
signal and y(t) represents the desired output signal.
We assume that x(t) is the general excitation signal shown in Fig. 14.27 (a). Also assume that x(t) 5 0 for
t , 0.
Now we see approximate x(t) by a series of rectangular pulses of uniform width Dt as shown in Fig.
14.27 (b). Thus,
x(t) 5 x0(t) 1 x1(t) 1 ... 1 xi(t) 1 ...
Fig. 14.27
(14.66)
630
Circuits and Networks
where xi(t) is a rectangular pulse that equals x(ti) between ti and ti 1 1 and is zero elsewhere. Note that the ith
pulse can be expressed in terms of step functions; that is
xi(t) 5 x(ti) [u(t – ti) – u(t – (ti 1 Dt))]
(14.67)
The next step in the approximation of x(t) is to
make Dt small enough that the ith component can be
approximated by an impulse function of strength x(ti) Dt.
Figure 14.27 (c) shows the impulse representation, with
the strength of each impulse shown in brackets beside
each arrow. The impulse representation of x(t) is
x(t) 5 x (t0) Dt d (t – t0) 1 x (t1) Dt d (t – t1) 1 ....
1 x (ti) Dt d (t – ti) 1 ....
(14.68)
Now when x(t) is represented by a series of impulse
functions, the response function y (t) consists of the sum
of a series of uniformly delayed impulse responses. The
strength of each response depends on the strength of the
impulse driving the circuit. For example, let’s assume that
the unit impulse response of the circuit contained within
the box in Fig. 14.26 is the exponential decay function
shown in Fig. 14.28 (a). Then the approximation of y (t) is
the sum of the impulse responses shown in Fig. 14.28 (b).
Fig. 14.28
y(t)
Analytically, the expression for y(t) is
5 x(t0) Dt h(t – t0) 1 x(t1) Dt h(t – t1)
1 x(t2) Dt h(t – t2) 1 ...
1 x(ti) Dt h(t – ti) 1 ...
(14.69)
As Dt → 0, the summation in Eq. (14.69) approaches a continuous integration, or
∞
∞
∑ x(ti ) h(t − ti ) ∆t → ∫ x(t) h(t − t)d t
i =0
(14.70)
0
Therefore,
∞
y(t ) = ∫ x( t) h(t − t) d t
(14.71)
0
If x(t) exists over all time, then the lower limit on Eq. (14.71) becomes – `, thus, in general
∞
y (t ) =
∫
x( t) h(t − t) d t
−∞
(14.72)
The integral relation between y(t), h(t) and x(t) is written in a shorthand notation
y(t) 5 h(t) * x(t) 5 x(t) * h(t)
(14.73)
Application of the Laplace Transform in Circuit Analysis
631
Thus, h(t) * x(t) is read as “h(t) is
convolved with x(t)” and implies that
∞
x(t ) * h(t ) =
∫
x( t) h(t − t) d t
(14.74)
−∞
The above integral gives the most
general relation for the convolution of two
functions. However, in our applications,
we can change the lower limit to zero
and the upper limit to t. Then the above
equation can be written as
t
y(t ) = ∫ x( t) h(t − t) d t
0
t
= ∫ h( t) x(t − t) d t
(14.75)
0
For physically realizable circuits, h(t)
is zero for t , 0. In other words, there can
be no impulse response before an impulse
is applied. We start measuring time at the
instant the excitation x(t) is turned on,
therefore x(t) 5 0 for t , 0.
A graphical interpretation of the
convolution integrals contained in
Eq. (14.75) is important in the use
of integral as a computational tool.
Consider the impulse response of our
circuit is the exponential decay function
shown in Fig. 14.29 (a) and the excitation
function has the waveform shown in Fig.
14.29 (b).
Replacing t with –t simply
Fig. 14.29
folds the excitation function over
the vertical axis and replacing
– l with t –l slides the folded function to the right. This folding and sliding
operation gives rise to the term convolution. At any specified value of t, the response
function y(t) is the area under the product function l(t) x(t – t) as shown in Fig. 14.29 (c). For t , 0, the
product l(t) x(t – t) is zero because h(t) is zero. For t . t, the product h(t) x(t – t) is zero because x(t – t)
is zero.
632
14.6
Circuits and Networks
THE TRANSFER FUNCTION AND THE STEADY-STATE SINUSOIDAL RESPONSE
LO 4
We use the transfer function to relate the steady-state response to the excitation source. First we assume that
x(t) 5 A cos (vt 1 f)
(14.76)
and we use the equation
Y(S ) 5 H(S ) X(S )
to find the steady-state solution of y(t).
To find the Laplace transform of x(t), we first write x(t) as
x(t) 5 A cos vt cos f – A sin vt sin f
(14.77)
(14.78)
from which
X (S ) =
=
A cos f S
S +v
2
2
−
A sin f v
S 2 + v2
A( S cos f − v sin f)
S 2 + v2
Substituting Eq. (14.79) into Eq. (14.77) gives the s-domain expression for the response
Y ( S ) = H ( s)
A( S cos f − v sin f)
S 2 + v2
(14.79)
(14.80)
By taking partial fractions
Y (S ) =
k1
k1*
+
S − jv S + jv
1 S terms generated by the poles of H(S )
(14.81)
In Eq. (14.81), the first two terms result from the complex conjugate poles of the deriving source. However,
the terms generated by the poles of H(s) do not contribute to the steady-state response of y(t), because all
these poles lie in the left half of the s-plane, consequently, the corresponding time-domain terms approach
zero as t increases.
Thus, the first two terms on the right-hand side of Eq. (14.81) determine the steady-state response. Now
K1 can be determined.
k1 =
=
H ( S ) A( S cos f − v sin f)
S + jv
S = jv
H ( jv) A( jv cos f − v sin f)
2 jv
A(cos f + j sin f) 1
= H ( jv) Ae jf
2
2
In general, H( jv) is a complex quantity, thus
= H ( j v)
(14.82)
H( jv) 5 |H( jv)| e ju(v)
(14.83)
where |H( jv)| is the magnitude, and phase angle is u(v) of the transfer function vary with the frequency v,
633
Application of the Laplace Transform in Circuit Analysis
the expression for K1 becomes
K1 =
A
|H ( jv)| e j[u(v)+f ]
2
(14.84)
We obtain the steady-state solution for y(t) by taking inverse transform of Eq. (14.81) ignoring the terms
generated by the poles of H(S ). Thus,
yss(t) 5 A|H( jv)| cos [vt 1 f 1 u(v)]
(14.85)
which indicates how to use the transfer function to find the steady-state sinusoidal response of a circuit.
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 4
A rectangular voltage pulse vi 5 [u(t) – u(t – 1)] V is
applied to the circuit shown in Fig. Q.1. Use convolution
to find v0.
rrr 14-4.2 Interchange the inductor and resistor in Problem 14.4.1
and again use the convolution integral to find v0.
rrr 14-4.3 The current source in the circuit shown is delivering
10 cos 4t A. Use the transfer function to compute the
steady-state expression for v0.
rrr 14-4.1
Fig. Q.1
Fig. Q.3
14.7
THE IMPULSE FUNCTION IN CIRCUIT ANALYSIS
Impulse functions occur in circuit analysis either because of a switching operation
or because a circuit is excited by an impulse source. The Laplace transform can be
used to predict the impulsive currents and voltages created during switching and
the response of a circuit to an impulsive source.
LO 5
14.7.1 Switching Operation
We use two different circuits to illustrate how an impulse function can be created with a switching operation:
a capacitor circuit and a series inductor circuit.
‰ Capacitor‰Circuit‰ In the circuit shown in Fig. 14.30, the capacitor C1 is charged to an initial voltage of V0 at the time the switch
is closed.
In the circuit, the initial charge on C2 is zero. Figure 14.31 shows
the s-domain equivalent circuit.
Fig. 14.30
634
Circuits and Networks
From Fig. 14.31,
I=
V0 /S
 1   1
 + 
R + 
 SC   SC
1
=
V0 /R
 1
S + 
 RC



2
(14.86)



e
Fig. 14.31
C1 C2
where the equivalent capacitance
is replaced by Ce.
C1 + C2
By taking inverse transform of Eq. (14.86), we obtain
i=
V0 −t / RCe
e
R
(14.87)
V 
which indicates that as R decreases, the initial current  0  increases and the time constant (RCe) decreases.
 R 
Thus, R gets smaller, the current starts from a larger initial value
and then dropped off more rapidly. Figure 14.32 shows these
characteristics of i.
The characteristics show, i is approaching an impulse function as
R approaching to zero because the initial value of i is approaching
infinity and time duration of i is approaching zero. We still have to
determine whether the area under the current function is independent
of R. Physically, the total area under the i versus t curve represents
the total charge transferred to C2 after the switch is closed. Thus,
Fig. 14.32
∞
V0 −t / RCe
e
dt = V0 Ce
R
0−
Area = q = ∫
(14.88)
which says that the total charge transferred to C2 is independent of time and equals V0Ce coulombs. Thus, as
R approaches zero, the current approaches an impulse strength V0Ce.
i → V0Ce d(t)
(14.89)
when R 5 0, a finite amount of charge is transferred to C2 instantaneously. When the switch is closed, the
voltage across C2 does not jump to V0 but its final value of
v2 =
C1V0
C1 + C2
(14.90)
If we set R equal to zero, the Laplace transform analysis will predict the impulsive current response.
Thus,
V0 /S
CCV
(14.91)
= 1 2 0 = CeV0
I=
 1   1  C1 + C2




 SC  +  SC 
1
2
The inverse transform of the above equation is
i 5 Ce V0 d(t)
(14.92)
Application of the Laplace Transform in Circuit Analysis
635
‰ Series‰Inductor‰Circuit‰ The circuit shown
in Fig. 14.33 illustrates a second switching operation that produces an impulsive response. The
problem is to find the time-domain expression for
v0– after the switch has been opened. Note that
opening the switch forces an instantaneous change
in the current of L2, which causes v0 to contain an
impulsive component.
Figure 14.34 shows the s-domain equivalent
with the switch open. The current in the 3 H
inductor at t 5 0 is 10 A, and the current in 2 H
inductor at t 5 0 is zero.
Fig. 14.33
Applying Kirchhoff’s current law, we get
V0
V − [(100 /S ) + 30]
+ 0
=0
2S + 15
3S + 10
(14.93)
Solving for V0 yields
V0 =
40( S + 7.5) 12( S + 7.5)
+
S ( S + 5)
S +5
(14.94)
Fig. 14.34
By taking partial fractions, we get
60
20
30
−
+ 12 +
S S +5
S +5
60
10
= 12 + +
S S +5
By taking inverse transform, we have
V0 =
v0 5 12 d(t) 1 (60 1 10e–5t) u(t) volts
(14.95)
(14.96)
Let us derive the expression for the current when t . 0. After the switch has been opened, the current in
L1 is the same as the current in L2. The current equation is
100 

+ 30
 S 
20
6
I=
=
+
S ( S + 5) S + 5
5S + 25
6
4
4
+
= −
S S +5 S +5
4
2
+
S S +5
By taking inverse transform gives
=
i 5 (4 1 2e–5t) u(t) A
(14.97)
(14.98)
Before the switch is opened, the current in L1 is 10 A, and the current in L2 is 0 A. We know that at
t 5 0, the current in L1 and in L2 is 6 A. Then, the current in L1 changes instantaneously from 10 A to 6 A,
636
Circuits and Networks
while the current in L2 changes instantaneously from
0 to 6 A. From this value of 6 A, the current decreases
exponentially to a final value of 4 A. Figure 14.35
shows these characteristics of i1 and i2.
Impulse‰Sources
Impulse functions can occur in sources as well as
responses. Such sources are called impulsive sources.
An impulse source driving a circuit imparts a finite
amount of energy into the system instantaneously. In
the circuit shown in Fig.
Fig. 14.35
14.36 (a), an impulsive
voltage source having a strength of V0 volt-seconds is applied to a series
connection of a resistor and an inductor. When the voltage source is applied, the
initial energy in the inductor is zero, therefore the initial current is zero. There is
no voltage drop across R. So the impulse voltage source appears directly across
L. An impulse voltage at the terminals of an inductor establishes an instantaneous
current. The current is
Fig. 14.36 (a)
t
1
i = ∫ V0 d( x ) dx
(14.99)
L 0
The integral of d(t) over any interval that includes zero is one; thus, we have
V
i(0) = 0 A
(14.100)
L
For an infinitesimal moment, the impulsive voltage source has stored in the inductor.
2
1 V 
1 V02
(14.101)
W = L  0  =
J
2  L 
2 L
V
The current 0 decays to zero in accordance with the natural response of the circuit, that is,
L
V
i = 0 e−t / t u(t )
(14.102)
L
L
where
t= .
R
When a circuit is driven by only an impulsive source, the total response
is completely defined by the natural response. The duration of the impulse
source is so infinitesimal that it does not contribute to any forced response.
We may also obtain Eq. (14.102) by direct application of the Laplace
transform method. Figure 14.36 (b) shows the s-domain equivalent of the
circuit in Fig. 14.36 (a).
The current I in the circuit is
V0
V /L
Fig. 14.36 (b)
I=
= 0
(14.103(a))
R + SL S + R /L
Taking inverse Laplace transform, we get
 R 
V0 − L  t V0 −t / t
e
= e u (t )
L
L
Thus, the Laplace transform method gives the correct solution for i $ 0.
i=
(14.103(b))
Application of the Laplace Transform in Circuit Analysis
637
Additional Solved Problems
PROBLEM 14.1
A 500 V resistor, a 16 mH inductor, and a 25 mF capacitor are connected in parallel. Express the admittance
of this parallel combination of elements as a rational function of S.
Solution‰ The circuit represented in s-domain of the above problem is shown in Fig. 14.37.
The admittance of terminals ab is
Y (S ) =
1
1
+
+ SC
R SL
(14.104)
Substituting the numerical values in the above
equation,
Fig. 14.37
Y (S ) =
1
1
+
+ S × 25×10−9
500 S ×16×10−3
(14.105)
Simplifying the above equation, we have
Y (S ) =
25×10−9 2
( S + 80, 000 S + 25×108 )
S
(14.106)
PROBLEM 14.2
The switch in the circuit shown has been in the position a for a long time. At t 5 0, the switch is thrown to
the position b. Find the current I as rational function of s. Find the time-domain expression for the current i.
Solution ‰ When the switch is at position for a long time, both the capacitors are charged to 100 V.
When the switch is at the position b, the s-domain circuit is shown in Fig. 14.39.
Fig. 14.38
Fig. 14.39
By applying Kirchhoff’s voltage law, we have
V1 V2
1
1
I+
I + I (5 K )
+ =
S
S
C1S
C2 S
(14.107)
638
Circuits and Networks

1
1
1
I
(V1 + V2 ) = 
+
+ 5×103 
−
6
−
6


S
S  0.2×10
0.8×10

I
I
1
100 = 
+ 5×103 
6
−


S
S  0.16 ×10
I=
0.02
S + 1250
(14.108)
By taking inverse transform, we get the time-domain expression for i
i 5 0.02 e21250t A.
(14.109)
PROBLEM 14.3
Obtain the current s-domain expression for the current IL in the circuit shown in Fig. 14.40. Also obtain the
time-domain expression for the inductor current. The switch is opened at t 5 0. Assume initial energy stored
in the circuit is zero.
Fig. 14.40
Solution
The s-domain equivalent circuit for the circuit shown in Fig. 14.40 is shown in Fig. 14.41.
Fig. 14.41
By applying Kirchhoff’s current law, we get
I
V V
+
= dc
R SL
S
(14.110)
I dc /C

1 
1
S 2 + 
S+
 RC 
LC
(14.111)
SCV +
V=
We know
IL =
V
SL
(14.112)
Application of the Laplace Transform in Circuit Analysis
639
Substituting Eq. (14.111) into Eq. (14.112), we get
IL =
I dc /LC
 2  1 

 S +  1 
S  S + 
 RC 
 LC 


(14.113)
Substituting the numerical values yields
IL =
384×105
S ( S 2 + 64000 S + 16×108 )
(14.114)
By taking partial fractions, we get
IL =
384 ×105
S ( S + 32000 − j 24000) ( S + 32000 + j 24000)
(14.115)
IL =
K1
K2
K 2*
+
+
S
S + 32000 − j 24000 S + 32000 − j 24000
(14.116)
The partial fraction coefficients are
384 ×105
K1 =
16 ×10
K2 =
8
= 24 ×10−3
384 ×105
(−32000 + j 24000) ( j 48000)
5 20 3 1023 ∠126.87°
(14.117)
Substituting the numerical values of K1 and K2 into Eq. (14.116) and inverse transforming the resulting
expression yields
iL 5 [24 1 40 e232,000 t cos (24000 t 1 126.87°)] mA
(14.118)
PROBLEM 14.4
Obtain the s-domain expression for the current IL in the circuit shown in Fig. 14.40 when the dc current
source is replaced by a sinusoidal current source ig 5 Im cos vt. Where Im 5 24 mA and v 5 40,000 rad/s.
Assume initial energy stored in the circuit is zero.
Solution ‰ The s-domain expression for the source current is
Ig =
SI m
s + v2
2
(14.119)
The voltage across the parallel elements is
V=
( I g /C ) s
 1 
1
s 2 + 
s+
 RC 
LC
(14.120)
640
Circuits and Networks
Substituting Eq. (14.119) into Eq. (14.120) results in
V=
( I m /C ) s 2

 1   1 

( s 2 + v2 )  s 2 + 
s +
 RC   LC 

(14.121)
from which
IL =
V
=
SL
( I m /LC ) s

 1   1 

( s 2 + v2 )  s 2 + 
s +
 RC   LC 


(14.122)
Substituting the numerical values of Im, v, R, L, and C in Eq. (14.122) gives
IL =
384×105 s
( s 2 + 16×108 ) ( s 2 + 64000 s + 16×108 )
(14.123)
By factoring the denominator, we get
IL =
384×105 s
( s 2 − jv) ( s + jv) ( s + a − jb) ( s + a + jb)
(14.124)
where v 5 40000, a 5 32000 and b 5 24000
By taking partial fractions, we get
IL =
K1
K1*
+
s − j 40000 s + j 40000
K2
K 2*
+
+
s + 32000 − j 24000 s + 32000 + j 24000
(14.125)
The coefficients K1 and K2 are
K1 =
384 ×105 ( j 40000)
( j80000) (32000 + j 16000) (32000 + j 64000)
5 7.5 3 10–3
–90°
(14.126)
384×10 (−32000 + j 24000)
(−32000 − j 16000) (−32000 + j 64000) ( j 48000)
5
K2 =
5 12.5 3 10–3
90°
(14.127)
Substituting the numerical values from (14.126) and (14.127) into (14.125) and inverse-transforming the
resulting expression yields
iL 5 [15 cos (40000 t – 90°) 1 25 e232000 t cos (24000 t 1 90°)] mA
iL 5 [15 sin 40000 t – 25 e232000 t sin 24000 t] mA
(14.128)
PROBLEM 14.5
Obtain the expression for i1 and i2 in the circuit shown in Fig. 14.42 when dc voltage source is applied
suddenly. Assume that the initial energy stored in the circuit is zero.
Application of the Laplace Transform in Circuit Analysis
641
Fig. 14.42
Solution ‰ Figure 14.43 shows the s-domain equivalent circuit for the circuit shown in Fig. 14.42.
Fig. 14.43
The two mesh current equations are
336
= ( 42 + 8.4 s) I1 − 42 I 2
s
0 5 – 42 I1 1 (90 1 10 s) I2
Using Cramer’s method to solve for I1 and I2, we get
D=
42 + 8.4 s
−42
−42
90 + 10 s
5 84
1 14 s 1 24)
5 84 (s 1 2) (s 1 12)
(14.129)
(14.130)
(s2
D1 =
D2 =
(14.131)
336 /s
−42
3360 ( s + 9)
=
0
90 + 10 s
s
42 + 8.4 s 336 /s
−42
0
=
14112
s
Based on Eqs (14.131) to (14.133),
D
40( s + 9)
I1 = 1 =
D
s( s + 2) ( s + 12)
D2
168
=
D
s( s + 2) ( s + 12)
Expanding I1 and I2 into a sum of partial fractions gives
I2 =
I1 =
15
14
1
−
−
s s + 2 s + 12
7 8.4
1.4
I2 = −
+
s s + 2 s + 12
(14.132)
(14.133)
(14.134)
(14.135)
(14.136)
(14.137)
642
Circuits and Networks
We obtain the expressions for i1 and i2 by inverse transforming Eq. (14.136) and (14.137) respectively
i1 5 (15 – 14 e22t – e212t) A
(14.138)
i2 5 (7 – 8.4 e22t 1 1.4 e212t) A
(14.139)
PROBLEM 14.6
Transform the circuit shown in Fig. 14.44 to the s-domain and
determine the Laplace impedance.
Solution‰ The transformed circuit for the above circuit is shown in
Fig. 14.45.
The parallel combination of inductor and capacitor is in series
with the resistor.
Fig. 14.44
4
s  
 5 




  4 
z = 3+
s �   = 3 +

4

  s 
s+
s
3s 2 + 4 s + 12
z=
s2 + 4
Fig. 14.45
(14.140)
PROBLEM 14.7
Determine the current i if the circuit is driven by a voltage
source as shown in Fig. 14.46. The initial value of the voltage
across the capacitor and the initial current through the inductor
are both zero.
The transformed circuit is as shown in Fig. 14.47.
Total Laplace impedance across the voltage source is
2
Z = 3+ s +
s
s 2 + 3s + 2
Z=
(14.141)
s
Thus, the current is
Solution
I=
I=
V
40 / ( s + 4)
= 2
Z ( s + 3s + 2) / s
40 s
( s + 4 ) ( s 2 + 3s + 2 )
Fig. 14.46
(14.142)
(14.143)
Fig. 14.47
By taking partial fractions,
I=
K
K1
K
+ 2 + 3
s +1 s + 2 s + 4
(14.144)
643
Application of the Laplace Transform in Circuit Analysis
The coefficients K1, K2, and K3 are
−40
K1 = I ( s + 1) s=−1 =
3
K 2 = I ( s + 2) s=−2 = 40
−80
K3 = I ×( s + 4) s=−4 =
3
Substituting the coefficients and taking inverse Laplace transform, we get
i = 40 e−2t −
40 −t 80 −4t
e − e
3
3
(14.145)
PROBLEM 14.8
Determine the current i for t $ 0 if initial current i(0) 5 1 for the
circuit shown in Fig. 14.48.
Solution‰ The s-domain circuit with series initial current in the
inductor is shown in Fig. 14.49.
Applying Kirchhoff’s voltage law results in
Fig. 14.48
10
− 4 I − 2 sI + 2 = 0
s
10 – 4sI – 2s2I 1 2s 5 0
Fig. 14.49
I=
10 + 2 s
2 s( s + 2)
I=
5+ s
s( s + 2)
(14.146)
(14.147)
Taking partial fractions
K
K
(14.148)
I= 1+ 2
s
s+2
The coefficients K1 and K2 are
5
−3
K1 = ; K 2 =
2
2
Substituting coefficients and taking inverse Laplace transform of Eq. (14.148) gives
5 3
i = − e−2 t
(14.149)
2 2
Alternatively, the inductor initial condition can be represented parallelly as shown in Fig. 14.50.
Fig. 14.50
644
Circuits and Networks
By inspection,
1
I2 =
s
By applying Kirchhoff’s voltage law to the mesh, we get
10
− 4 I1 − 2 sI1 + 2 sI 2 = 0
s
1
From the figure, I1 = I and I 2 =
s
(14.150)
1
10
− 4 I − 2 sI + 2 s   = 0
 s 
s
s+5
I=
s( s + 2)
(14.151)
Taking partial fractions and inverse Laplace transform, we get
5 3
i = − e−2 t
2 2
(14.152)
PROBLEM 14.9
Determine the current i for t $ 0 if Vc (0) 5 4 V for the circuit shown in
Fig. 14.51.
The transformed s-domain circuit is shown in Fig. 14.52.
Application of Kirchhoff’s voltage law gives
Solution‰
8
20
4
− 4 I −   I − = 0

 5
s
s
4
I=
s+2
Fig. 14.51
(14.153)
Taking inverse Laplace transform gives
Fig. 14.52
i 5 4 e22t
(14.154)
Alternatively, the initial condition can be represented as shown
in Fig. 14.53.
Fig. 14.53
645
Application of the Laplace Transform in Circuit Analysis
By inspection, we have
−1
I2 =
2
Applying Kirchhoff’s voltage law to the mesh 1 results in
 8  −1
20
8
− 4 I1 − I1 +    = 0
 s  2 
s
S
(14.155)
−1
2
 8  −1
20
8
− 4 I − I +    = 0
 s  2 
s
s
4
I=
s+2
Taking inverse transform, we get
i 5 4 e22t
Because I1 = I and I 2 =
(14.156)
(14.157)
PROBLEM 14.10
Convert the current source in Fig. 14.54 to a voltage source in the
s-domain.
Solution‰ Converting the circuit in Fig. 14.54 into the voltage source
results in the circuit shown in Fig. 14.55.
Fig. 14.54
Fig. 14.55
PROBLEM 14.11
Convert the voltage source in Fig. 14.56 to a current source in the
s-domain.
Fig. 14.56
646
Circuits and Networks
Solution‰ Converting Fig. 14.56 into a current source in the s-domain results in the circuit shown in Fig. 14.57.
Fig. 14.57
PROBLEM 14.12
Determine v0 for the circuit shown in Fig. 14.58.
Solution‰ The circuit in Fig. 14.58 in the s-domain is as
shown in Fig. 14.59.
Fig. 14.58
Total impedance in the circuit
Zeq 5 Z1 1 Z2 1 Z3
(14.158)


Z1
V
V0 = 
 Z1 + Z 2 + Z3 



 1 










4




2
=





2
s + 4



2s + 4 + 






s








=
=
(14.159)
Fig. 14.59
s
( s + 4) ( s + 2 s + 1)
2
s
( s + 4) ( s + 1) 2
(14.160)
Taking partial fractions,
V0 =
K1
( s + 1)
2
+
K
K2
+ 3
s +1 s + 4
The coefficients K1, K2, and K3 are
K1 =
−1
4
−4
, K 2 = , K3 =
3
9
9
(14.161)
647
Application of the Laplace Transform in Circuit Analysis
Thus,
V0 =
(−1/ 3)
( s + 1) 2
+
4 / 9 (−4 / 9)
+
s +1
s+4
Taking inverse transform on both sides,
1
4
4
v0 = t e−t + e−t − e−4t
3
9
9
(14.162)
PROBLEM 14.13
Determine i1, i2, V and V1, for the circuit in Fig. 14.60.
Solution‰ The transformed circuit in the s-domain is
shown in Fig. 14.61.
Fig. 14.60
Applying current division to the circuit in the s-domain, we get
 s 3s 
Zeq = 4 储  + 
 4 4 
(14.163)
 Zeq  6 
I1 = 
 
 Z1 + Z 2  s 


Fig. 14.61
=
24
s( s + 4 )
(14.164)
By taking partial fraction expansion,
I1 =
K1
K
+ 2
s
s+4
K1 = 6; K 2 = −6
6 (−6)
I1 = +
s s+4
(14.165)
Taking inverse transform, we get
i1 5 6 – 6 e24t
(14.166)
648
Circuits and Networks
From Kirchhoff’s current law,
6
= I 2 + I1
s
= I2 +
(14.167)
24
s( s + 4 )
6
24
I2 = −
s s( s + 4 )
I2 =
6
s+4
(14.168)
We know
V = 4 I2 =
24
s+4
(14.169)


 3s 
V1 = 
 I1
 

4



 
 24 



 3s 

=

 





+
4
(
)
s
s

4





18
V1 =
s+4
Taking inverse transforms,
(14.170)

 6 

−4 t

i2 = L −1 { I 2 } = L −1 

= 6e


 s + 4


(14.171)

 24 

−4 t

v = L −1 {V } =L −1 
 = 24 e


s
+
4




(14.172)

 18 

−4

v1 = L −1 {V1 } = L −1 

 = 18 e


 s + 4


(14.173)
PROBLEM 14.14
Determine the voltage v for the circuit shown in Fig. 14.62.
Fig. 14.62
Application of the Laplace Transform in Circuit Analysis
649
Solution‰ The circuit in Fig. 14.62 is transformed into s-domain as shown in Fig. 14.63.
Fig. 14.63
By using mesh analysis, the current I2 in the circuit is
1 12 4
−
s s s
−1
2
I2 =
1
1+
−1
s
−1 1 + 1 + s
(14.174)
 1 8
2 1 +  +  
 s   s 
2 s + 10
= 2
I2 =
 1 
1 +  ( 2 + s) −1 s + 2 s + 2
 s 
(14.175)
1+
The voltage across the 1 V resistor is
V = R I2 =
2 s + 10
(14.176)
s + 2s + 2
2
The above equation can be written as
V=
2s + 2 + 8
s2 + 2s + 2

 s +1 





1


= 2
 2
 2
 + 8







 s + 2s + 2 
 s + 2s + 2 





 s +1 



1



= 2
+
8




2
2




+
+
(
s
)
(
s
+
1
)
+
1
1
1








(14.177)
Taking inverse Laplace transform on both sides,
 s + 1 


v = 2L −1 

 + 8L
2


(
s
)
+
1
+
1




v 5 2 e2t cos t 1 8 e2t sin t


1


2


(
s
+
1
)
+
1



−1 

(14.178)
650
Circuits and Networks
PROBLEM 14.15
Determine the voltage v for the circuit in Fig. 14.64. Assume vc(0) 5 0.
Fig. 14.64
Solution‰ The circuit in Fig. 14.64 is transformed into the s-domain resulting in the circuit in Fig. 14.65.
Fig. 14.65
Replacing the Laplace impedance for R and C with Laplace admittance, we get Fig. 14.66
Fig. 14.66
By using nodal analysis, we get
V1 =
1
s
4
s
s
4
−s
4
1+
−s
4
s
4
−s
4
2+
2+
s
4
(14.179)
Application of the Laplace Transform in Circuit Analysis
 1 
 s 
  4  −s 
 
−   
2 +  


 s  4 
 s   4 





V1 =


  s 
  −s  −s 
 s 


1 +  

2 +  
 −   








4
4






  4  4 
651
(14.180)
 5   8 
  s +  
 3   3 
=

8
s  s + 

3
Taking partial fraction expansion,
K
K
V1 = 1 + 2
8
s
s+
3
The coefficients K1 and K2 are
2
K1 = 1; K 2 =
3
1 2/3
V1 = +
8
s
s+
3
Taking the inverse Laplace transform of each side of the equation results in










(
/
)
1
2
3
−1
−1 

L {V1 } = L  +





8
s



s
+








3



(14.181)
(14.182)
(14.183)

 1 
1 2


v1 = L −1   + L −1 


 s  3



 s + 8 / 3

8
2 − t
v1 = 1 + e 3
3
and because v 5 v1
 8 
2 − t
v = 1 + e 3
3
(14.184)
PROBLEM 14.16
Determine the voltage v for the circuit shown in
Fig. 14.67 using Thevenin’s theorem.
Fig. 14.67
652
Circuits and Networks
Solution‰ We have to find out the open-circuit voltage as shown in Fig. 14.68.
Fig. 14.68
Applying the superposition method results in the circuits in Figs 14.69 and 14.70, where V9T and V0T are the
1
 4
contributions to VT from the Laplace transformed sources   and   respectively.
 s 
 s 
Fig. 14.69
Fig. 14.70
From Fig. 14.69, the open- circuit voltage is
 1  4 1  8 + s
VT′ = IZ =   +  =
 s  s 2  2 s 2
(14.185)
From Fig. 14.70, the open-circuit voltage is
 4  1  2
VT′′ = IZ =    =
 s  2  s
(14.186)
because no current flows through the capacitor.
From the superposition method,
VT = VT′ + VT′′
=
=
VT =
8+ s
2s
2
+
2
s
(14.187)
8 + s + 4s
2s2
5s + 8
2s2
(14.188)
653
Application of the Laplace Transform in Circuit Analysis
Replacing both current sources by opens, as
required, Thevenin’s theorem to determine the Thevenin
impedance results in Fig. 14.71.
The impedance seen into the terminals ab
ZT =
4 1
+
s 2
s +8
2s
and the Thevenin equivalent circuit for terminals a–b is shown in Fig. 14.72.
=
Fig. 14.71
(14.189)
If the 1 V resistor is reconnected across terminals ab then V can be determined in Fig. 14.73.
Fig. 14.72
Fig. 14.73









 5s + 8 
1



V =



2




+
8
s
s
2






+
1



2s 







 5s + 8 

 2s 


=



2 



3
s
+
8




 2s 


5s + 8
V=
s(3s + 8)
(14.190)
(14.191)
The inverse Laplace transform of V is
8
 2  − t
v = 1 +   e  3
 3 
(14.192)
PROBLEM 14.17
Determine the voltage V for the circuit shown in Fig. 14.74,
using Norton’s theorem.
Solution‰ The application of Norton’s theorem in the
s-domain requires the removal of the 1 V resistor as
shown in Fig. 14.75 and the determination of resulting
short-circuited current.
Fig. 14.74
654
Circuits and Networks
Fig. 14.75
Applying the superposition method results in the circuits in Figs 14.76 and 14.77, where IN9 and IN0 are the
1
 4
contributions to IN from the Laplace transformed current sources   and   respectively
 s 
 s 
By inspection of the circuit in Fig. 14.76,
I N′ =
1
s
(14.193)
Fig. 14.76
Fig. 14.77
Applying current division to the circuit in Fig. 14.77 results in
 1 



  
 




 2 
4
 4

I N′′ = 
=


 








1
4
+
s
s
8


 

  +  









 2   s  

(14.194)
From the superposition method,
IN 5 I9N 1 IN0
1
4
= +
s s +8
IN =
5s + 8
s( s + 8)
(14.195)
Because Thevenin and Norton impedances are equal,
ZN =
s +8
2s
(14.196)
The Norton equivalent circuit for terminals a–b is as shown in Fig. 14.78.
If the 1 V resistor is reconnected across terminals ab, the voltage V can be determined in the circuit shown
in Fig. 14.79.
Application of the Laplace Transform in Circuit Analysis
655
Fig. 14.79
Fig. 14.78
 s + 8 

1
 2 s { }
 s + 8 
Z = 
储
(1)
=

 2 s 
s +8
1+
2s
s +8
Z=
3s + 8
(14.197)
(14.198)
We know V 5 ZI


 s +8 
 5s + 8 




=









 3s + 8 

 s( s + 8) 

5s + 8
V=
s(3s + 8)
Taking inverse Laplace transform of V, we get
(14.199)
8
 2  − t
v = 1 +   e  3
 3 
(14.200)
PROBLEM 14.18
The initial charge on the capacitor in the circuit shown in Fig. 14.80 is zero.
(a) Find the s-domain Thevenin equivalent circuit with respect to terminals a and b.
(b) Find the s-domain expression for the current that the circuit delivers to a load consisting of a 1 H inductor
in series with a 2 V resistor.
Fig. 14.80
Solution‰ First, we have to find out the Thevenin's equivalent circuit from the s-domain circuit shown in Fig.
14.81.
Thevenin's voltage across terminals ab is
Vab 5 Vx 1 0.2 Vx – I(1)
(14.201)
656
Circuits and Networks
Fig. 14.81
By applying Kirchhoff’s voltage law, we can determine the current I

1 

0.2Vx = I 1 +
 0.5s 
0.1sVx
I=
(1 + 0.5s)
Since no current is passing through the 5 V resistor,
20
The voltage Vx =
s
Substituting Vx and I in Eq. (14.201), we get
(14.202)
(14.203)
 20 
0.1s  20 
Vab = 1.2   −
 s  1 + 0.5s  s 
(14.204)
20  s + 2.4 


s  s + 2 
The Thevenin’s impedance after short-circuiting the voltage sources is shown in Fig. 14.82.
Vab =
(14.205)
 1 



 储 (1) 
Z ab = 

+ 5





 0.5s 

Z ab =
5( s + 2.4)
s+2
(14.206)
The Thevenin’s equivalent circuit is shown in Fig. 14.83.
Fig. 14.82
Fig. 14.83
657
Application of the Laplace Transform in Circuit Analysis
The current I in the circuit of Fig. 14.83 is
20  s + 2.4 


s  s + 2 
I=
5( s + 2.4)
+2+s
s+2
=
I=
(14.207)

20 
( s + 2.4)


2
s  5s + 12 + s + 4 s + 4 
(14.208)
20  s + 2.4 


s  s 2 + 9 s + 16 
(14.209)
PROBLEM 14.19
The voltage source vg drives the circuit shown in
Fig. 14.84. The response signal is the voltage across
the capacitor vo. Calculate the numerical expression
for the transfer function.
Solution‰ The s-domain equivalent circuit is shown in
Fig. 14.84
Fig. 14.85.
Fig. 14.85
By definition, transfer function is the ratio vo/vg.
By applying Kirchhoff’s current law, we get
V0 −Vg
1000
+
V0
Vs
+ 0 =0
250 + 0.05s 106
Solving for V0 yields
1000( s + 5000)Vg
V0 = 2
s + 6000 s + 25×106
Hence, the transfer function is
V
1000( s + 5000)
H ( s) = 0 = 2
Vg
s + 6000 s + 25×106
(14.210)
(14.211)
(14.212)
658
Circuits and Networks
PROBLEM 14.20
The circuit shown in Fig. 14.86 is driven by a voltage
source whose voltage increases linearly with time,
namely, v0 5 50 t u(t).
(a) Use the transfer function to find v0.
(b) Identify the transient component of the
response.
(c) Identify the steady-state component of the
response.
Fig. 14.86
Solution‰ From the previous example
H (S ) =
1000( s + 5000)
s + 6000 s + 25×106
2
(14.213)
The transform of the driving voltage is 50/s2, therefore, the s-domain expression for the output voltage is
1000( s + 5000)
50
V0 = 2
(14.214)
6
( s + 6000 s + 25×10 ) s 2
The partial fraction expansion of V0 is
V0 =
K
K1
K1*
K
+
+ 2+ 3
s + 3000 − j 4000 s + 3000 + j 4000 s 2
s
(14.215)
The involves of coefficients are
K1 5 5 5 3 10– 4
79.70°
K1* 5 5 5 3 10– 4
–79.70°
K2 5 10
K3 5 – 4 3 10– 4
(a) The time domain expression for v0 is
v0 5 [10 5 3 10 – 4 e–3000t cos (4000 t 1 79.70°)
1 10 t – 4 3 10 – 4] V
(14.216)
(b) The transient component of v0 is
10 5 3 10 – 4 e–3000t cos(4000 t 1 79.70°) V
(14.217)
(c) The steady-state component of the response is
(10 t – 4 3 10 – 4) V
PROBLEM 14.21
The excitation voltage vi for the circuit shown in Fig. 14.87 is shown in Fig. 14.88.
(a) Use convolution integral to find v0.
(b) Plot v0 over the range of 0 # t # 15 s.
(14.218)
Application of the Laplace Transform in Circuit Analysis
Fig. 14.87
659
Fig. 14.88
Solution‰ The first step in using the convolution integral is to find the unit impulse response of the circuit.
Obtain the expression for V0 from the s-domain equivalent of the circuit in Fig. 14.87.
V0 =
Vi
(1)
s +1
(14.219)
When vi is a unit impulse function d(t),
vo 5 h(t)
5 e–t u(t)
(14.220)
from which
h(t) 5 e–l u(t)
(14.221)
The impulse response and the folded excitation function is shown in Fig. 14.89.
Sliding the folded excitation function to the right requires breaking the integration into intervals:
0 # t # 5; 5 # t # 10; and 10 # t # `. The breaks in the excitation function at 0.5, and 10s dictate these
break points. Figure 14.90 shows the positioning of the folded excitation for each of these intervals. The
analytical expression for vi in the time interval 0 # t # 5 is
vi 5 4t; 0 # t # 5s
(14.222)
(a)
(b)
Fig. 14.89
Hence, the analytical expression for the folded excitation function in the interval t – 5 # t # t is
vi (t – t) 5 4 (t – t), t – 5 # t # t
We can now set up the three integral expression for v0.
(14.223)
660
Circuits and Networks
Fig. 14.90
For 0 # t # 5s,
t
v0 = ∫ 4(t − t)e−t d t
0
5 4(e – t 1 t – 1) V
For 5 # t # 10s,
t −5
v0 =
∫
(14.224)
t
20e−t d t + ∫ 4(t − t)e−t d t
t −5
0
5 4 (5 1
e–t
–
e – (t – 5)) V
(14.225)
661
Application of the Laplace Transform in Circuit Analysis
For 0 # t # ` s,
t −5
v0 =
∫
t −10
t
20e−t d t + ∫ 4(t − t)e−t d t
t −5
5 4 (e – t – e – (t – 5) 1 5e – (t – 10)) V
(14.226)
The results are computed for v0 and tabulated in Table 14.1.
The voltage response is shown graphically in Fig. 14.91.
Fig. 14.91
Table 14.1
Numerical Values of v0 (t)
t
v0
t
v0
t
v0
1
1.47
6
18.54
11
7.35
2
4.54
7
19.56
12
2.70
3
8.20
8
19.8
13
0.99
4
12.07
9
19.93
14
0.37
5
16.03
10
19.97
15
0.13
PROBLEM 14.22
For the circuit shown in Fig. 14.92, the sinusiodal
source voltage is vg 5 120 cos (5000 t 1 30°) V. Find
the steady-state expression for V0.
Solution‰ From Problem 14.19,
H (S ) =
1000( s + 5000)
s + 6000 s + 25×106
2
(14.227)
The frequency of the voltage source is 5000 rad/s;
Hence, we evaluate H(S ) at H(j5000).
H ( j 5000) =
=
1000(5000 + j 5000)
−25×10−6 + j 5000(6000) + 25×106
2
1 + j1 1− j1
=
=
∠− 45°
6
6
j6
Then the steady-state voltage is
v0 ss =
120 2
cos (5000t + 30° − 45°)
6
= 20 2 cos (5000t −15°) V.
Fig. 14.92
662
Circuits and Networks
PSpice Problems
PROBLEM 14.1
The circuit shown in Fig. 14.93 consists of series RL elements. The sine
wave is applied to the circuit when the switch ‘S’ is closed at t 5 0. Find
i (t) using PSpice.
f =
25
= 3.979
2
RL TRANSIENT WITH SINE WAVE INPUT
VS 1
0
SIN(0 5 3.979 0)
R
1
2
10
L
2
0
5
.TRAN 0.1 1
.PROBE
.PLOT TRAN I(L)
.END
**** TRANSIENT ANALYSIS TEMPERATURE 5 27.000 DEG C
****************************************************************
TIME
I(L)
0.000E 1 00 0.000E 1 00
1.000E – 01 6.417E – 02
2.000E – 01 1.264E – 02
3.000E – 01 1.102E – 02
4.000E – 01 4.784E – 02
5.000E – 01 –2.371E – 02
6.000E – 01 4.252E – 02
7.000E – 01 –1.548E – 03
8.000E – 01 – 4.937E – 03
9.000E – 01 3.820E – 02
1.000E 1 00 –3.371E – 02
Fig. 14.93
PROBLEM 14.2
Using PSpice, for the circuit shown in Fig. 14.94,
find the voltage across the 0.5 V resistor when switch
s is opened at t 5 0. Assume there is no charge on
the capacitor and no current in the inductor before
switching.
Fig. 14.94
663
Application of the Laplace Transform in Circuit Analysis
PWL( 0 5 0.1N 5 10 5)
0.5
1
1
Fig. 14.95
I1
0
1
R
1
0
L1 0
1
C1 0
1
.TRAN 0.05 0.5
.PROBE
664
Circuits and Networks
.PRINT TRAN V(R)
.END
**** TRANSIENT ANALYSIS TEMPERATURE 5 27.000 DEG C
****************************************************************
TIME
V(R)
0.000E 1 00 0.000E 1 00
5.000E – 02 –8.794E – 18
1.000E – 01 –8.794E – 18
1.500E – 01 –8.794E – 18
2.000E – 01 –8.794E – 18
2.500E – 01 –8.794E – 18
3.000E – 01 –8.794E – 18
3.500E – 01 –8.794E – 18
4.000E – 01 –8.794E – 18
4.500E – 01 –8.794E – 18
5.000E – 01 –8.281E – 18
Answers to Practice Problems
14-2.1
2000 (S 1 50000)2/(S 2 1 80000S 1 25 3 108)
14-2.3
(a)
I 5 40/(S 2 1 12S 1 1)
(b) i 5 50e–0.6t sin 0.8t A
(c) V 5 16S/(S 2 1 1.2S 1 1)
14-2.5
(d)
v 5 200 e–0.6t cos(0.8t 1 36.87°) V
(a)
100  −8t
(100 / 3)e−2t − 
e V
 3 
(b)
50 −2t 50 −8t
e − e
V
3
3
(c) 50 e–2t – 50 e–8t V
14-2.8
(a)
14-2.10 (a)
(e)
2+
10 −t
e cos(3t −126.87°) V
3
80 V
16 V
(b) 20 V
(f) 4 V
14-3.1
H(S ) 5 10(S 1 2)/S 2 1 2S 1 10
14-4.1
v 5 (1 – e–t) V
V 5 (e – 1)e–t V
14-4.3
44.7 cos (4t – 63.43°) V
(b)
10.54e–t cos (3t – 18.43°) V
(c) 0 V
(g) 20 V
0#t#1
1#t#`
(d)
32 d(t) mA
665
Application of the Laplace Transform in Circuit Analysis
Objective-Type Questions
rrr 14.1
An inductor in the s-domain consists of
(a)
(b)
(c)
(d)
rrr 14.2
current source in series with an inductor
voltage source in parallel with an inductor
voltage source of LI0 in series with an inductor
current source I0/s in series with an inductor
A capacitor in the s-domain consists of
(a) current source CV0 in parallel with capcitor
(b) current source in series with capacitor
(c) voltage source
V0 in parallel with capacitor
s
(d) voltage source CV0 in parallel with capacitor
rrr 14.3
The current in the circuit when the switch is closed at t 5 0.
(a) 10 e–100 t
(c) 0.1
rrr 14.4
e–1000 t
(b) 0.01 e–1000 t
(d) 10 e–0.1 t
The initial voltage across the capacitor when the switch s is
opened at t 5 0.
Fig. 14.96
Fig. 14.97
(a) zero
rrr 14.5
(b) C .
I dc
s
(c)
1
I dc
CS
(d) CS ( I dc )
Thevenins equivalent circuit across terminals ab.
(a) The voltage source 10 V with (20 1 2s) impedance in
series
(b) The voltage source 10 V in parallel with (20 1 2s) V
impedance
10
(c) The voltage source
in series with an impedance of
s
(20 1 2s) V
10
(d) The voltage source
in series with an impedance
s
of 22 V
rrr 14.6
Fig. 14.98
The transfer function of multiple independent sources can
easily be obtained by
(a) superposition theorem
(b) Thevenin’s theorem
(c) Norton’s theorem
(d) reciprocity
666
rrr 14.7
Circuits and Networks
In the circuit shown in Fig. 14.99, the current is defined as
the response signal then the transfer function
(a)
(b)
(c)
(d)
rrr 14.8
10−6 s
I
10−12 s 2 + s + 1
s
s2 + s +1
s
Fig. 14.99
s2 +1
s
s +1
The circuit is driven by an unit impulse source then the response equals to
(a) transfer function
(b) one
rrr 14.9
(c) zero
(d) inverse of transfer function
If the input of a circuit is represented by series of impulse functions, the response consists of
(a)
(b)
(c)
(d)
sum of the series of uniformly delayed impulse responses
sum of the series of responses
one
zero
rrr 14.10 For physically realizable circuit, impulse response is
(a) zero for t < 0
(b) zero for t > 0
(c) one for t < 0
(d) infinite for t > 0
rrr 14.11 The instantaneous current in an inductor when an impulse voltage V0 applied to the terminals of an inductor
(a)
zero
(b) unity
(c)
For interactive quiz with answers,
scan the QR code given here
OR
visit
http://qrcode.flipick.com/index.php/272
V0
L
(d)
V0
d(t )
L
15
LEARNING OBJECTIVES
15.1
CONCEPT OF COMPLEX FREQUENCY
The solution of differential equations for networks is of the form
Sn t
i(t) 5 K e
n
LO 1
(15.1)
where Sn is a complex number which is a root of the characteristic equation
and may, therefore, be expressed as
Sn 5 sn 1 jvn
(15.2)
The complex number consists of two parts, the real part sn and the imaginary part vn. The real part of the
complex frequency sn is neper frequency, while the imaginary part vn is the radian frequency. The neper
frequency has dimensions of neper per second. In the time-domain equations, vn is in the form of sin vnt or
cos vnt. The radian frequency vn is expressed in radians/second and is related to the frequency fn in cycles/
sec or the periodic time T (in seconds) by the relation.
v n = 2p f n =
2p
T
(15.3)
From Eq. (15.2), we see that the real part sn and the imaginary part vn must have identical dimensions.
2p
Radian frequency vn is
has dimensions (time)21. Therefore, the dimensions of sn must also be (time)–1 or
T
668
Circuits and Networks
the unit of sn must be “something per unit time”. Since sn appears as an exponential factor,
I 5 I0 esnt
(15.4)
1 I 
s n = ln  
t  I 0 
(15.5)
Such that
It is fact that “something per unit time” should be nondimensional logarithmic unit. The usual unit for
the natural logarithm is the neper, making the dimensions for s the neper per second. The complex quantity
(15.6)
is defined as the complex frequency. Thus, complex frequency consists of a real part sn called the neper frequency
and an imaginary part vn is called radian frequency (or real frequency).
15.2
PHYSICAL INTERPRETATION OF COMPLEX FREQUENCY
LO 1
The complex frequency appears in the exponential form e Snt. Let us consider the physical significance of complex
frequency and a number of special cases for the values of Sn.
‰ Case‰(i)‰ Let Sn 5 sn 1 jo having zero radian frequency. The exponential function becomes
Kn eSnt 5 Knesnt
(15.7)
The above exponential function increases exponentially for sn . 0
and decreases exponentially for sn , 0. For sn 5 0, the exponential
function reduces to Kn and it is a time-invariant resulting current i(t)
which is a dc current. Figure 15.1 shows the variation of exponential
term Kn esnt with time t for the cases of sn . 0, sn , 0 and sn 5 0.
Fig. 15.1
‰ Case‰ (ii)‰ Let Sn 5 0 6 jvn having radian frequency and zero
neper frequency. The exponential becomes
i(t) 5 Kn e 6jvnt
5 Kn (cos vnt 6 j sin vnt)
(15.8)
The exponential e6jvnt may be interpreted in terms of the physical model of a rotating phasor of unit
length. A positive sign of exponential e6jvnt implies counter- clockwise or positive rotation, while a negative
sign e–jvnt implies clockwise or negative rotation.
For positive or counter-clockwise rotation, the real part of e1jvnt or the projection on the real axis equals
cos vnt, while the imaginary part of e1jvnt or the projection on the imaginary axis equals sin vnt (Fig. 15.2).
The variation of exponential function e1jvnt with time is thus sinusoidal and hence constitutes the case of
sinusoidal steady state.
‰ Case‰(iii)‰ Let Sn 5 sn 1 jvn is the general case and the frequency is complex and the exponential is
given by
i(t) 5 Kn eSnt 5 Kn e(sn 1 jvn)t
i(t) 5 Kn esnt ? e jvnt
(15.9)
669
S-Domain Analysis
Fig. 15.2
Equation (15.9) shows that with complex frequency, the time variation of response i(t) is the product of
the response for Sn 5 sn 1 jo and the response for Sn 5 0 1 jvn. The response esnt for the case of neper
frequency alone, Sn 5 sn 1 jo is an exponentially increasing or decaying function. The response ejvnt for
the case of radian frequency alone Sn 5 0 1 jvn may be represented by a rotating phasor. The product e snt
? e jvnt may then be visualized as a rotating phasor whose magnitude is not constant at unity but changes
continuously with time. Such phasors are illustrated in Fig. 15.3.
Fig. 15.3 (a)
Fig. 15.3 (b)
The real and imaginary projections of this phasor are
Re(eSnt) 5 esnt cos vnt
and
(15.10)
Im(eSnt) 5 esnt sin vnt
Consider the projections of this rotating phasor on the real and imaginary axes for the two cases. These
projections for the case sn , 0 are known as a damped sinusoid and for sn . 0, the increasing oscillations are
shown in Figs 15.4 (a) and (b) respectively.
From the above discussion, it is clear that the imaginary part of complex frequency governs sinusoidal
oscillations and the real part of complex frequency governs the exponential decay or rise.
The roles of two types of frequency are similar even though the variations caused by them are different.
This is the justification of unifying the two concepts under the name of complex frequency.
670
Circuits and Networks
(a)
(b)
Fig. 15.4
15.3
TRANSFORM IMPEDANCE AND TRANSFORM CIRCUITS
In this section, we determine the transform impedance and admittance
representations for each of the elements and initial condition sources.
LO 2
15.3.1 Resistance
For a resistance, the voltage and current are related in the time domain by Ohm’s law.
1
VR (t ) = RiR (t ) or iR (t ) = GVR (t ); G =
R
The corresponding transform equations are
(15.11)
VR(S ) 5 RIR(S )
IR(S ) 5 GVR(S )
(15.12)
The ratio of transform voltage VR(S ) to the transform current IR(S ) is defined as the transform impedance
of the resistor, expressed as
(15.13)
Similarly, the reciprocal of this ratio is the transform admittance for the resistor, expressed as
(15.14)
S-Domain Analysis
Fig. 15.5 (a)
671
From the above results, we can say that the resistor is frequency
insensitive to the complex frequency.
Figure 15.5 (a) shows a network representing resistor R current
iR(t) and voltage VR(t) in time domain. Figure 15.5 (b) gives the
network representation of the same resistor and also transform
current IR(S ) and voltage VR(S ).
Fig. 15.5 (b)
15.3.2 Inductance
For inductance, the time-domain relation between the current in inductance iL(t) and the voltage VL(t) across
it is expressed as
v L (t ) = L
diL (t )
dt
t
and iL (t ) =
1
∫ vL (t ) dt
L -∞
(15.15)
The equivalent transform equation for the voltage expression is
VL(S ) 5 L [SIL(S ) – iL(01)]
(15.16)
which, on rearranging, results
LS IL(S ) 5 VL(S ) 1 LiL(01)
(15.17)
In Eqs (15.15) and (15.16), VL(S ) is the transform of the applied voltage vL(t) and LiL(01) is the transform
voltage caused by the initial current iL(01) present in the inductor at time t 5 01.
Considering the sum of the transform voltage and the initial current voltage as V1(S ), we have the transform
impedance for the inductor.
Z L (S ) =
V1 ( S )
= SL
I L (S )
(15.18)
Figure 15.6 (a) shows the time-domain network representation of the inductor
L, the current iL(t) through it, and the applied voltage vL(t ). Figure 15.6 (b) gives
the transform representation of same inductor in terms of impedance using Eq.
(15.16).
The transform equation for the current expression of Eq. (15.17) is
Fig. 15.6 (a)
0+




v
(
t
)
dt


L
∫
V ( S ) -∞
1

I L (S ) =  L
+
 S
 L
S
(15.19)
0+
But
Fig. 15.6 (b)
∫
-∞
vL (t ) dt = LiL (0+)
(15.20)
672
Circuits and Networks
Hence, Eq. (15.19) becomes
1 VL ( S ) iL (0+)
⋅
+
L
S
S
iL (0+)
1
VL ( S ) = I L ( S ) LS
S
I L (S ) =
or
(15.21)
(15.22)
In the above equation, iL(01)/S is the transform caused by the initial current iL(01) in the inductor.
i (0+)
Let I1 ( S ) = I L ( S ) - L
(15.23)
S
Then Eq. (15.22) becomes
1
VL ( S ) = I1 ( S )
LS
(15.24)
where I1(S ) is the total transform current through the inductor L. The transform admittance becomes
YL ( S ) =
I1 ( S )
1
=
VL ( S ) LS
Fig. 15.7 (a)
Fig. 15.7 (b)
(15.25)
Figure 15.7 (a) shows the time-domain
representation of inductor L with initial current iL(01).
Figure 15.7 (b) shows equivalent transform circuit thus
1
contains an admittance of value
and equivalent
LS
transform current source.
15.3.3 Capacitance
For capacitance, the time-domain relation between voltage and current is expressed as
dv (t )
ic (t ) = c c
dt
t
and
1
 c (t ) = ∫ ic (t ) dt
C -∞
(15.26)
The equivalent transform equation for the voltage expression is
VC ( S ) =
1
C
 I C ( S ) q(0+) 


+
S 
 S
(15.27)
q(0+)
= vC (0+) is the initial voltage across the capacitor.
C
The above equation becomes
where
v (0+)
1
I C ( S ) = VC ( S ) - C
CS
S
(15.28)
Considering the total transform voltage across the capacitor as V1(S ).
V1 ( S ) = VC ( S ) -
vC (0+)
S
(15.29)
673
S-Domain Analysis
Then, Eq. (15.28) becomes
1
(15.30)
I C ( S ) = V1 ( S )
CS
The transform impedance of the capacitor is the ratio of transform voltage V1(S ) to the transform current
IC(S ) and is
V (S )
1
(15.31)
ZC ( S ) = 1
=
I C ( S ) CS
Figure 15.8 (a) shows the time-domain representation
of capacitor C with initial voltage VC (01) across it.
Voltage V1(S ) includes the initial voltage VC (01). Figure
15.8 (b) gives the transform representation of the same
capacitor in terms of transform impedance.
Considering the current expression, the transform
equation corresponding to Eq. (15.26) is
Fig. 15.8 (a)
Fig. 15.8 (b)
IC (S ) 5 C [SVC (S ) – nC (01)]
(15.32)
On rearranging,
CSVC (S ) 5 IC (S ) 1 CVC (01)
(15.33)
Considering the transform current through YC (S ) as I1(S ), Eq. (15.32) may be put as
CSVC (S ) 5 I1(S )
(15.34)
Then the transform admittance of the capacitor C is the ratio of transform current I1(S ) to transform
voltage VC (S ), expressed as
YC ( S ) =
I1 ( S )
= CS
VC ( S )
(15.35)
Figure 15.9 (a) shows the time-domain representation
of the capacitor C with initial voltage VC (01) across it.
Figure 15.9 (b) gives the transform representation of the
same capacitor in terms of admittance.
Fig. 15.9 (a)
Fig. 15.9 (b)
Frequently Asked Questions linked to LO 2*
rrr 15-2.1
L1= 1H
Obtain the transfer impedance for the circuit shown in
Fig. Q.1.
[BPUT 2008]
C1 1F
L2
1H
1 ohm
Fig. Q.1
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
R
V2
674
15.4
Circuits and Networks
SERIES AND PARALLEL COMBINATIONS OF ELEMENTS
In general, any network can be connected in series, parallel, and series-parallel
LO 3
combinations. Consider the series combination of elements shown in Fig. 15.10.
Assume that all
initial conditions are
zero, i.e. the current in all inductors is zero and that
the initial voltage of all capacitors is also zero. By
application of Kirchhoff’s voltage law, we get
Fig. 15.10
n(t) 5 nR 1 nR 1 … 1 nL 1 nL 1 … 1 nC 1 nC 1 …
1
2
1
2
1
(15.36)
2
Taking transform for the above equation, we get
V(S ) 5 VR (S ) 1 … 1 VL (S ) 1 … 1 VC (S ) 1 …
1
1
(15.37)
1
Dividing the equation by I(S ), the transform current through the series circuit, we get
Z(S ) 5 ZR (S ) 1 … 1 ZL (S ) 1 … 1 ZC (S )
1
or
1
(15.38)
1
n
Z (S ) = ∑ Z K (S )
(15.39)
k =1
where n is the total number of elements of all
kinds in series.
Figure 15.11 shows the transform
representation of the series circuit and
represents Eq. (15.38).
Consider a parallel combination of resistors,
inductors, and capacitors as shown in Fig. 15.12. Here, we
assume that the inductors have zero initial currents and
capacitors have zero initial voltages. Let v(t) be the common
voltage applied to all the elements in the circuit.
Applying Kirchhoff’s current law to the above circuit
yields
Fig. 15.11
Fig. 15.12
i(t) 5 iG (t) 1 iG (t) 1 … 1 iL (t)
1
2
1
1 iL (t) 1 … 1 iC (t) 1 iC (t) 1 …
2
1
2
(15.40)
and the corresponding transform equation is
I(S ) 5 IG (S ) 1 IG (S ) 1 … 1 IL (S ) 1 IL (S ) 1 … 1 IC1(S )
1
2
1
2
1 IC (S ) 1 …
(15.41)
2
If this equation is divided by V(S ), we get the transform admittance which is the ratio of the current
transform to the voltage transform and is
Y(S ) 5 YG (S) 1 YG (S) 1 … 1 YL (S) 1 YL (S) 1 …1 YC (S )
1
2
1 YC (S ) 1 …
2
1
2
1
(15.42)
675
S-Domain Analysis
or
n
Y ( S ) = ∑ YK ( S )
(15.43)
k =1
where n is the total number of all kinds of elements in parallel.
Figure 15.13 gives the transform representation of the parallel network and represents Eq. (15.42).
For a series-parallel combination, rules for the combination of impedance and of admittance can be
used to reduce a network to a single equivalent impedance or admittance.
Fig. 15.13
EXAMPLE 15.1
In the circuit shown, the switch K is moved from the position 1 to the position 2 at time t 5 0. At time t 5 02,
the current through inductor L is I0 and the voltage across capacitor is V0. Find the transform current I(S).
Solution‰ The inductor has an initial current of I0. It is represented by a transform impedance LS in series with
a voltage source LI0 as shown in Fig. 15.15. The capacitor has an initial voltage V0 across it. It is represented
V0
1
with an initial voltage
. The transform circuit derived from the circuit
CS
S
of Fig. 15.14 is shown in Fig. 15.15.
by a transform impedance of
Fig. 15.14
Fig. 15.15
The current I(S ) is given as the total transform voltage in the circuit divided by the total transform
impedance. Then
V
V1 ( S ) + LI 0 - 0
V (S )
S = SV1 ( S ) + SLI 0 -V0
=
I (S ) =
(15.44)
1
1
Z (S )
LS 2 + RS +
R + LS +
CS
C
EXAMPLE 15.2
The network shown in Fig. 15.16 is a parallel combination
of L, R and C connected across a current source I. At time
t 5 0 –, the current through inductor L is I0 and the voltage
across capacitor C is V0. At time t 5 01, the current source
I1(t) is connected to the parallel RLC circuit. Find the
transform voltage V(S).
Fig. 15.16
676
Circuits and Networks
Solution‰ Figure 15.17 gives the transform network
corresponding to the given network with the switch
K moved to the position 2.
From the above transform circuit, the transform
voltage V(S ) may be obtained by taking the ratio
of the total transform current to the total transform
admittance.
The total transform current in the network is given by
I
I ( S ) = I1 ( S ) - 0 + CV0
S
Total transform admittance is given by
1
Y (S ) = G +
+ CS
LS
Hence, the transform voltage is given by
I
I1 ( S ) + CV0 - 0
I (S )
S
V (S ) =
=
1
Y (S )
+ CS
G+
LS
Fig. 15.17
EXAMPLE 15.3
Obtain the transform impedance of the network shown in Fig. 15.18.
Solution‰ The transform network of Fig. 15.18 is shown in Fig. 15.19.
The admittance of the last two elements is the parallel combination.
Y1(S ) 5 4 1 S
1
Therefore, impedance is Z1 ( S ) =
S +4
Series combination of last elements
S + 4 + 2S
1
1
3S + 4
Z2 (S ) =
+
=
=
2S S + 4
2S ( S + 4)
2S ( S + 4)
Parallel combination of elements
Y2 ( S ) =
Fig. 15.19
Fig. 15.18
1 2S ( S + 4) (3S + 4) + 4 s( S + 4) 4S 2 + 19S + 4
+
=
=
2
3S + 4
6S + 8
6S + 8
1
6s + 8
Hence, the impedence Z ( S ) =
=
Y2 ( S ) 4 S 2 + 19S + 4
EXAMPLE 15.4
In the given network in Fig. 15.20, the switch S is opened at t 5 0,
the steady state having established previously. With the switch S
open, draw the transform network for analysis on the loop basis
representing all elements and all initial conditions. Write transform
equation for current in the loop.
Fig. 15.20
S-Domain Analysis
677
Solution‰ Under steady-state conditions, the capacitor is open-circuited and the
10
= 5 A.
2
The voltage across the capacitor is V0 5 10 V. Hence, the corresponding transform
network is shown in Fig. 15.21.
10
5+
5( S + 2)
V (S )
S
Hence, I ( S ) =
=
=
1 S 2 + 6S + 1
Z (S )
2+S +4+
S
inductor is short-circuited. The current through the inductor is i0 =
Fig. 15.21
15.5
TERMINAL PAIRS OR PORTS
LO 3
Consider an arbitrary network made up of passive elements. It can be represented by a rectangular box shown
in Fig. 15.22.
For the network shown in Fig. 15.22 (a), only one
voltage and one current exist and only one network
function is defined. It constitutes one pair of terminals
called a port. Generally, a driving source is connected
to the pair of terminals. For the two-terminal pair
network shown in Fig. 15.22 (b), two currents and two
Fig. 15.22
voltages must exist. Normally, in Fig. 15.22 (b), 1-19
and 2-29 are called ports. Hence, it is a called two-port network. If the driving source is connected across 1-19, the
load is connected across 2-29. Otherwise, if the source is connected across 2-29, the output is taken across 1-19.
15.6
NETWORK FUNCTIONS FOR ONE-PORT AND TWO-PORT NETWORKS
For a one-port network, the driving-point impedance or impedance of the
LO 4
network is defined as
V ( s)
Z ( s) =
(15.45)
I ( s)
The reciprocal of the impedance function is the driving-point admittance function, and is denoted by Y(s).
For the two-port network without internal sources, the driving-point impedance function at the port 1-19 is
the ratio of the transform voltage at port 1-19 to the transform current at the same port.
V ( s)
Z11 ( s) = 1
(15.46)
I1 ( s)
Similarly, the driving-point impedance at the port 2-29 is the ratio of transform voltage at port 2-29 to the
transform current at the same port.
V ( s)
(15.47)
Z 22 ( s) = 2
I 2 ( s)
For the two-port network, the driving-point admittance is defined as the ratio of the transform current at
any port to the transform voltage at the same port.
Therefore, Y11 ( s) =
I1 ( s)
V1 ( s)
(15.48)
678
Circuits and Networks
I 2 ( s)
, which is the driving-point admittance.
V2 ( s)
The four other network functions are called transfer functions. These functions give the relation between
voltage or current at one port to the voltage or current at the other port as shown hereunder.
or Y22 ( s) =
‰ Voltage-Transfer‰Ratio‰ This is the ratio of voltage transform at one port to the voltage transform at
the other port, and is denoted by G(s).
V ( s)
G21 ( s) = 2
V1 ( s)
and G12 ( s) =
V1 ( s)
V2 ( s)
(15.49)
‰ Current-Transfer‰ Ratio‰ This is the ratio of current transform at one port to current transform at
other port, and is denoted by a(s).
I ( s)
a12 ( s) = 1
I 2 ( s)
and a 21 ( s) =
I 2 ( s)
I1 ( s)
(15.50)
‰ Transfer‰Impedance‰ It is defined as the ratio of voltage transform at one port to the current transform at the other port, and is denoted by Z(s).
and
Z 21 ( s) =
V2 ( s)
I1 ( s)
Z12 ( s) =
V1 ( s)
I 2 ( s)
(15.51)
‰ Transfer‰Admittance‰ It is defined as the ratio of current transform at one port to the voltage transform at the other port, and is denoted by Y(s).
I ( s)
Y21 ( s) = 2
V1 ( s)
I ( s)
and Y12 ( s) = 1
(15.52)
V2 ( s)
The above network functions are found by forming the system of equations using node or mesh analysis,
and taking the transforms of equations by setting the initial conditions to zero and solving for ratio of the
response to excitation.
EXAMPLE 15.5
For the network shown in Fig. 15.23, obtain the drivingpoint impedance.
Solution‰ Applying Kirchhoff’s law at the port 1-19,
Z (S ) =
V (S )
I (S )
Fig. 15.23
679
S-Domain Analysis
where V(S ) is applied at the port 1-19 and I(S ) is the current flowing through the network. Then
V (S )
1
Z (S ) =
= 2+S +
I (S )
S
Z (S ) =
S 2 + 2S + 1
S
EXAMPLE 15.6
For the network shown in Fig. 15.24, obtain the transfer
functions G21(S) and Z21(S) and the driving-point impedance
Z11(S).
Solution‰ Applying Kirchhoff’s law,
V1(S ) 5 2I1(S ) 1 2SI1(S )
V2(S ) 5 I1(S ) 3 2S
Fig. 15.24
Hence,
V2 ( S )
S
2S
=
=
V1 ( S ) 2 + 2S S + 1
V (S )
= 2S
Z 21 ( S ) = 2
I1 ( S )
V (S )
Z11 ( S ) = 1
= 2( S + 1)
I1 ( S )
G21 ( S ) =
EXAMPLE 15.7
For the network shown in Fig. 15.25, obtain the transfer
functions G21(S ), Z21(S), and driving-point impedance Z11(S).
Solution‰ From the circuit, the parallel combination of
resistance and capacitance can be combined into equivalent
impedance.
1
2
Zeq ( S ) =
=
1 4S + 1
2S +
2
Applying Kirchhoff’s laws, we have
V2(S ) 5 2I1(S )
 2

and V1 ( S ) = I1 ( S ) 
+ 2
 4 S + 1

 8S + 4 

= I1 ( S ) 
 4 S + 1 
Fig. 15.25
680
Circuits and Networks
The transfer functions
V (S )
2 I1 ( S )
8S + 2
=
G21 ( S ) = 2
=


V1 ( S )  8S + 4 
8S + 4
 I (S )

4 S + 1  1
V (S )
Z 21 ( S ) = 2
=2
I1 ( S )
The driving-point function is
Z11 ( S ) =
V1 ( S ) 8S + 4
=
I1 ( S ) 4 S + 1
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 4
Calculate the driving-point functions for the following circuit shown in Fig.Q.1.
rrr 15-4.2 Find the transfer function G21(s) for the two-port network shown in Fig. Q.2.
rrr 15-4.1
Fig. Q.1
rrr 15-4.3
Fig. Q.2
Determine the driving-point impedance Zd of the network shown in Fig. Q.3.
Fig. Q.3
Frequently Asked Questions linked to LO 4*
I s I s
rrr
V s
0.5S
10 W
rrr 15-4.2
State and explain the initial and final-value theorems.
[GTU Dec. 2012]
rrr 15-4.3 Compute the driving-point impedance for the network
shown in Fig. Q.3.
[PTU 2011-12]
rrr
V1(S)
5
8
10 W
I1(S)
I2(S)
Fig. Q.1
Zs
*Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
Zs
s
V2(S)
681
S-Domain Analysis
2F
2F
R3
L
R1
C
Z(S)
R1
R2
R2
L2
L1
C1
C3
Z(S)
Fig. Q.3
rrr 15-4.5
rrr 15-4.6
Fig. Q.4
[PU 2010]
Find the driving-point function of the network shown in Fig. Q.5.
For the network shown in the Fig. Q.6, determine the transfer impedance.
1
5s
[RGTU June 2014]
s
I
20 ‡
10 ‡
3s
V (s)
I
5H
2H
Z(S)
½s
1
Fig. Q.5
rrr 15-4.7
1/4 ‡
1‡
Y(S)
Fig. Q. 6
Calculate the transfer impedance of given circuit shown in Fig. Q.7.
[RTU Feb. 2011]
V s
I s
[RTU Feb. 2011]
rrr
CC
R1
R2
I(t)
v(t)
L1
Fig. Q.8
Fig. Q.7
15.7
POLES AND ZEROS OF NETWORK FUNCTIONS
In general, the network function N(s) may be written as
a s n + a1s n-1 + … + an-1s + an
P ( s)
N ( s) =
= 0m
Q( s) b0 s + b1s m-1 + … + bm-1s + bm
LO 5
(15.53)
where a0, a1,..., an and b0, b1, ..., bm are the coefficients of the polynomials P(s) and Q(s); they are real
and positive for a passive network. If the numerator and denominator of polynomial N(s) are factorised, the
network function can be written as
a ( s - z1 )( s - z2 ) …( s - zn )
P ( s)
N ( s) =
= 0
(15.54)
Q( s) b0 ( s - p1 )( s - p2 ) …( s - pm )
where z1, z2, ..., zn are the n roots for P(s) 5 0
and p1, p2, ..., pm are the m roots for Q(s) 5 0
682
Circuits and Networks
and a0/b0 5 H is a constant called the scale factor.
z1, z2,..., zn in the transfer function are called zeros, and are denoted by 0. Similarly, p1, p2,..., pm are called
poles, and are denoted by 3. The network function N(s) becomes zero when s is equal to any one of the zeros.
N(s) becomes infinite when s is equal to any one of the poles. The network function is completely defined
by its poles and zeros. If the poles or zeros are not repeated, then the function is said to have simple poles
or simple zeros. If the poles or zeros are repeated, then the function is said to have multiple poles, multiple
zeros. When n . m, then (n – m) zeros are at s 5 `, and for m . n, (m – n) poles are at s 5 `.
Consider the network function
N ( s) =
(15.55)
that has double zeros at s 5 –1 and a zero at s 5 –5; and three
finite poles at s 5 –2, s 5 –3 1 j2, and s 5 –3 – j2 as shown
in Fig. 15.26.
The network function is said to be stable when the real parts
of the poles and zeros are negative. Otherwise, the poles and
zeros must lie within the left half of the s-plane.
Fig. 15.26
15.8
( s + 1) 2 ( s + 5)
( s + 2)( s + 3 + j 2)( s + 3 - j 2)
SIGNIFICANCE OF POLES AND ZEROS
LO 5
Poles and zeros are critical frequencies. At poles, the network function become infinite, while at zeros, the
network function becomes zero. At other complex frequencies, the network function has a finite non-zero
value.
Poles and zeros provide useful information in network functions. Consider the following cases.
‰ Driving‰Point‰Impedance
V (S )
(15.56)
I (S )
A pole of Z(S ) implies zero current for a finite voltage which means an open circuit. A zero of Z(S )
implies no voltage for a finite current or a short circuit.
1
Consider Z ( S ) =
(15.57)
CS
The above function has a pole at S 5 0 and zero at S 5 `.
Therefore, the above function represented by capacitor acts an open circuit at pole frequency and acts
as short circuit at zero frequency.
Z (S ) =
‰ Driving‰Point‰Admittance
Y (S ) =
I (S )
V (S )
(15.58)
A pole of Y(S ) implies zero voltage for a finite value of current which gives a short circuit. A zero of
Y(S ) implies zero current for a finite value of voltage which gives an open circuit.
‰ Voltage‰Transform‰Ratio
G21 ( S ) =
V2 ( S )
V1 ( S )
V2(S ) 5 G21(S ) V1(S )
(15.59)
683
S-Domain Analysis
To obtain output voltage, we require the product of input and transfer function. The expression for G21(S ).
V1(S ) is obtained in the form of a ratio of polynomials in S.
By making use of partial fractions, we can obtain a pole of either G21 (S ) or V1(S ) and no repeated
roots.
n
m
A
A
G21 ( S )V1 ( S ) = ∑
+∑
(15.60)
S
aj
S
a
i
i=1
j =1
where n and m are the number of poles of G21(S ) and V1(S ) respectively.
The frequencies ai from the natural complex frequencies corresponding to free oscillations and depend
on the network function G21(S ). While frequencies aj constitute the complex frequencies corresponding
to the forced oscillations and depend on the driving force V1(S ). From the above discussion, we can say
that the poles determine the time variation of the response whereas the zeros determine the magnitude
response.
‰Other‰Network‰Functions‰ Significance of poles and zeros in other transfer functions is the same
as discussed above. In each of the cases, poles determine the time-domain behaviour and zeros determine the magnitude of each of the terms of the response.
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 5
For the given network function, draw the pole zero diagram and, hence, obtain the time-domain
response. Verify the result analytically.
5( s + 5)
V ( s) =
( s + 2) ( s + 7)
rrr 15-5.2 Obtain the pole zero configuration of the impedance function of
the network shown in Fig.Q.2.
rrr 15-5.3 Draw the pole zero diagram for the given network function and
hence, obtain v(t).
4s
Fig. Q.2
V ( s) =
( s + 2) ( s + 3)
rrr 15-5.1
rrr 15-5.4
For the network shown in Fig. Q.4, determine the following transfer functions (a) G21(s), (b)
Y21(s), and (c) a21(s).
rrr 15-5.5
For the network shown in Fig. Q.5, determine the following functions (a) Z11(s), (b) Y11(s), (c)
G21(s), and (d) a21(s).
Fig. Q.4
Fig. Q.5
684
Circuits and Networks
Frequently Asked Questions linked to LO 5
r
r 15-5.2
15.9
1.
Explain the significance of poles and zeros.
PROPERTIES OF DRIVING POINT FUNCTIONS
The driving-point function is a ratio of polynomials in S. Polynomials
LO 6
are obtained from the transform impedances of the elements and
their combinations.
Let P ( S ) = a0 S n + a1S n-1 + … + an-1 S + an 
(15.61)

and Q( S ) = b0 S m + b1S m-1 + … + bm-1 S + bm 
be the numerator and the denominator polynomials respectively. The above equations can be factorised
and therefore written as
P ( S ) = ( S - Z1 ) ( S - Z 2 ) …( S - Z n )
(15.62)

Q( S ) = ( S - P1 ) ( S - P2 ) …( S - Pm ) 
The driving-point function N(S ) may be written as
(15.63)
The quantities Z1, Z2 Zn are called zeros of N(S ) as N(Z1) 5 N(Z2) 5 ... N(Zn) 5 0.
The quantities P1, P2
Pm are called poles of N(S ) as N(P1) 5 N(P2) 5
N(Pm) 5 `.
That is, pole is that finite value of S for which N(S ) becomes infinity.
If the zeros and poles are not repeated then the poles or zeros are said to be distinct or simple.
A zero or a pole is said to be of multiplicity ‘r’ if (S – Z )r or (S – P)r is a factor of P(S ) or Q(S ).
1
A function N(S ) is said to have a pole (or zero) at infinity, if the function N   has a pole (or zero)
S 
at S 5 0.
Consider the function
S +1
N (S ) =
( S + 2)( S + 4)
1
+1
 1 
S ( S + 1)
S
N   =
=
S  1


 + 2 1 + 4 (1 + 2S )(1 + 4 S )
 S

 S
i.e.
(15.64)
(15.65)
1
N   has a zero at S 5 0
S 
N(S ) has a zero at S 5 `
From the above example, we say that the number of zeros including zeros at ` equals the number
of poles including poles at `.
S-Domain Analysis
685
2.
(a) N(S ) be a driving-point impedance, i.e. Z(S )
V (S )
Z (S ) =
(15.66)
I (S )
A zero of N(S ) is a zero of V(S ), it signifies a short circuit. Similarly, a pole of Z(S ) is a zero of
I(S ). The poles of Z(S ) are those frequencies corresponding to open circuit conditions.
(b) Consider a driving-point admittance function
I (S )
Y (S ) =
(15.67)
V (S )
A zero of Y(S ) means a zero of I(S ), i.e. the open-circuit condition and a pole of Y(S ) means a
zero of V(S ) signifies a short circuit.
3.
Since all the elements in the circuit are real positive quantities the coefficients a0, a1, a2, ... an and b0,
b1, b2 ... bm are real and positive. Therefore, any zeros or poles, if complex, must occur in conjugate
pairs.
The real parts of all zeros and poles must be negative or zero. Consider a pole ‘P’ of N(S ), i.e. (S – P)
is a factor of the denominator of N(S ). Using partial fractions, we know that this gives rise to a term
A
of the form
whose inverse Laplace transform contains the term ept. The real part of ept tends
S-P
to zero as t tends to infinity if the real part P is negative. Therefore, for a finite input the response is
finite as t tends to infinity if the real part of P is negative. A network function whose response is finite
for all t, for a given finite input is said to be stable. Thus, a driving-point impedance Z(S ) is stable if
all the poles lie in the negative half of the S-plane.
1
Since Y ( S ) =
, poles of Y(S ) are zeros of Z(S ). Therefore Y(S ) is stable if all the zeros of
Z (S )
Z(S ) also lie in the negative half of the S-plane. Thus, the real parts of all zeros and poles of a driving
point function must be negative or zero.
Poles or zeros lying on the jv-axis must be simple. Consider a pole ‘P’ lying on the jv-axis. If it is not
simple then in the time response of the function of which it is a pole contains the term tk ejvt which
tends to infinity as t tends to infinite. Therefore, the function becomes unstable. Since zeros of one
function will be poles of the other, therefore, the zeros of driving point function should also satisfy
this condition.
4.
5.
6.
The degree of P(S ) and Q(S ) may differ by zero or one only.
At very high frequencies, the term a0 Sn dominates over the other terms in the numerator and the
term b0 Sm dominates over other terms in the denominator.
a
Lt N ( S ) = Lt 0 S n-m
(15.68)
S →∞
S →∞ b
0
Consider the network elements R, L, C, and M. R is independent of frequency.
a0
∴ if n 5 m, then the function behaves as a resistance R =
at high frequency. The impedance LS
b0
of an inductor increases linearly with the complex frequency S and, therefore, is an open circuit at
S 5 `. Thus if n 5 m 1 1, the function N(S ) behaves as an inductance as S approaches infinity. A
capacitor is a short circuit at infinite frequencies. Thus, N(S ) behaves as a capacitance if m 5 n 1 16.
Now, consider the driving-point impedance Z(S ). Z(S ) will behave as an inductor as LS increases
1
with increasing S while
decreases with increasing S and, therefore, the impedance of an inductance
CS
686
Circuits and Networks
dominates over the capacitive impedance. If inductors are not present in the circuit, then R dominates
1
over
as S tends to infinity. Thus, Z(S ) 5 LS or R as S tends infinity.
CS
∴ n – m 5 0 or 1
If N(S ) is a driving-point impedance.
1
On the other hand, the admittance of an inductance
tends to zero as S tends to infinity.
LS
Similarly, the admittance of a capacitance CS tends to infinity as S tends to infinity. Therefore, CS
1
dominates over
as S tends to infinity. If the network does not contain capacitors then the
LS
1
resistance R dominates over
at higher frequencies.
LS
∴ If N(S ) is a driving-point admittance function, m – n 5 0 or 1.
Therefore, |n – m| 5 0 or 1
7.
The lowest degree terms in P(S ) and Q(S ) may differ in degree by zero or one only.
As S approaches zero, the higher power of S tends to zero faster than S, therefore N(S ) can be
approximated by
N (S ) =
an-1S + an
bm-n S + bm
(15.69)
The impedance of an inductance ‘LS’ approaches zero as S tends to zero while that of a capacitor
1
1
approaches infinity as S tends to zero.
dominates over LS as S tends to zero. Therefore, for
CS
CS
Z(S ), the capacitance dominates over an inductance as S tends to zero. If the network does not contain
8.
capacitors then R dominates over LS and S tends to zero. Thus, the network can be replaced by R or
1
1 1
if Z(S ) is of interest. Similarly, for Y ( S ); ,
dominate over CS as S tends to zero. Therefore,
R LS
CS
for purposes of Y(S ), the network is just a conductance or an inductance. Thus, the network is just one
inductor or one capacitance or one resistance as S tends to zero.
K3
∴ N(S ) is of the form K1 or K2 S or
where K1, K2, K3 are constants.
S
Hence, the lowest degree of P(S ) and Q(S ) can differ at most in one degree.
P(S ) and Q(S ) cannot have missing terms unless all even or all odd degree terms are absent.
We know that
P(S ) 5 a0 Sn 1 a1 Sn – 1 1 ... 1 ai 1 1 Sn –i – 1 1 ai Sn – i. 1 ... 1 an – 1 S 1 an
and Q(S ) 5 b0 Sm 1 b1 Sm – 1 1 ... 1 bi 1 1 Sm – i 21 1 bi Sm – i
1 ... 1 bm – 1 S 1 bm
(15.70)
The above requirement means that for any i, ai or bi cannot be zero unless aj 5 0 or bj 5 0
for all i $ j. The only exception to this rule is when all even or all odd powers of S are missing. To
understand this, consider the network under study contains only elements like R, L, C, M whose
1
transfer impedances are R, LS,
, MS respectively.
CS
Also, R, L, C, M are positive quantities. A combination of RL or RC will give rise to a term of

b
b  aS + b
the form (aS 1 b) or a + . Since a +  =
means (aS 1 b) in the numerator and S in the

S
S
S
denominator. Similarly, R, L, C give rise terms of the form (aS 2 1 bS 1 C ) and a combination of only L
S-Domain Analysis
687
and C gives rise to a term of the form (aS 2 1 b). Therefore, when two such factors are multiplied, since
all the coefficients in each term are positive, in the expansion of the product no term can be zero. If all the
terms are of the form (aS 2 1 b), then their product contains only even powers of S. If this is multiplied
by S, the resulting function contains only odd powers of S. Given a ratio of polynomials N(S ), these
properties can therefore be used to find out if N(S ) represents a driving point function of a network.
15.10
1.
2.
3.
4.
PROPERTIES OF TRANSFER FUNCTIONS
LO 6
The transfer function is a ratio of polynomials in S.
The coefficients of P(S ), the numerators polynomial and of Q(S ), the denominator polynomial must
be real. Therefore, all poles and zeros, if complex, must occur in conjugate pairs.
The real parts of all poles must be negative and any pole on the jv-axis must be simple. As in the case
of driving-point functions, this follows from the stability considerations.
Since poles of the transfer function are zeros of Q(S ), it follows that the zeroes of Q(S ) must lie in the
negative half plane and any zero lying on the jv-axis must be simple.
Let P1, P2, ... Pm be the zeros of Q(S )
Then Q(S ) 5 K ? (S – P1) (S – P2) (S – P3) ... (S – Pm)
5.
Since all poles have negative real parts and complex poles occur in conjugate pairs, the product
of these factors contains all powers of S whose
coefficients are positive. Therefore, Q(S ) does
not have missing terms unless all even or all odd
powers are missing. Since there are no restrictions
on the zeroes of the transfer function, P(S ) can
have missing terms. Also coefficients of powers of
S in P(S ) can be negative.
For G(S ) and a(S ), the degree of the numerator
polynomial P(S ) is less than or equal to the degree
of Q(S ).
To prove this, we use the fact that a two-port network can be represented by an equivalent T(star)
or P (delta) shown in Fig. 15.27.
Let a source of known voltage V1(S ) be applied to a Tnetwork of the port 119. Let a source of known current I1(S ) be applied to the p-network at 111 and
Fig. 15.27
221 be short-circuited, then assuming I2(S ) 5 0.
G(S ) =
V2 ( S )
Z2 (S )
=
V1 ( S ) Z1 ( S ) + Z 2 ( S )
and a( S ) =
I 2 (S )
Y2 ( S )
=
I1 ( S ) Y1 ( S ) + Y2 ( S )
(15.71)
(15.72)
Since Z1(S ), Z2(S ), Z3(S ); Y1(S ), Y2(S ), and Y3(S ) can be thought off as the driving point functions of some
one ports, they have to satisfy the properties of driving point immittance functions.
688
Circuits and Networks
Since Z1(S ) and Z2(S ) are ratio of polynomials,
Let
Z1 ( S ) = K
Z2 (S ) =
( S + a1 )( S + a 2 ) …( S + a n1 )
( S + 1 )( S + 2 ) …( S + m1 )
K 2 ( S + r1 )( S + r2 ) …( S + rn 2 )
( S + d1 )( S + d2 ) …( S + dm 2 )
(15.73)
(15.74)
Substituting these expression in G(S ),
G(S ) =
=
Z2 (S )
Z1 ( S ) + Z 2 ( S )
(15.75)
K1 ( S + a1 )( S + a 2 ) …( S + a n1 )( S + d1 ) …( S + dm 2 )
( S + a1 )( S + a 2 ) …( S + a n1 )( S + d1 ) …( S + dm 2 )( S + 1 )
( S + 2 )( S + 3 ) …( S + n 2 )( S + 1 ) …( S + m1 )
Let P(S ) denote the numerator polynomial of G(S ) and Q(S ), the denominator polynomial of
G(S ).
Then degree of P(S ) 5 n1 1 m2 and degree of Q(S ) 5 n1 1 m2 or n2 1 m1, whichever is greater.
Thus, if n1 1 m2 . n2 1 m1, the degree of P(S ) equals the degree of Q(S ).
If n1 1 m2 , n2 1 m1, degree of Q(S ) 5 n2 1 m1 and the degree of P(S ) is less than the degree
of Q(S ).
Similarly, assuming Y1(S ) and Y2(S ) as ratios of polynomials and substituting those expressions
in a(S ), it can be shown that the degree of the numerator of a(S ) is less than or equal to the degree of
the denominator.
(f ) The degree of the numerator polynomial of Z21(S ) or Y21(S ) is less than or equal to the degree of the
denominator polynomial plus one.
Referring to the T and p equivalent networks of two-port network shown in Fig. 15.27,
Z 21 ( S ) =
V2 ( S )
I1 ( S )
=
I 2 =0
Z 2 ( S ) I1 ( S )
= Z2 (S )
I1 ( S )
(15.76)
I 2 (S )
-V ( S ) I1 ( S )
= 2
= -Y2 ( S )
(15.77)
V1 ( S ) V =0
I1 ( S )
2
Thus, the highest degree of the numerator of Z21(S ) equals the highest degree of the numerator of Z2(S ).
But as Z2(S ) is a driving-point impedance, the highest degree of the numerator of Z2(S ) is the degree of
denominator plus one. Therefore, the highest degree of the numerator of Z12(S ) is the degree of its denominator
plus one. Similarly, since Y2(S ) is a driving-point admittance, the highest degree of the numerator or Y21(S ),
which is also the numerator of Y2(S ) is equal to the degree of the denominator plus one.
and Y21 ( S ) =
15.11
NECESSARY CONDITIONS FOR A DRIVING POINT FUNCTION
LO 6
The restrictions on pole and zero locations in the driving-point function with common factors in P(s) and Q(s)
cancelled are listed below.
1. The coefficients in the polynomials P(s) and Q(s) of network function N(s) 5 P(s)/Q(s) must be real
and positive.
689
S-Domain Analysis
2.
3.
4.
5.
6.
Complex or imaginary poles and zeros must occur in conjugate pairs.
(a) The real parts of all poles and zeros must be zero, or negative.
(b) If the real part is zero, then the pole and zero must be simple.
The polynomials P(s) and Q(s) may not have any missing terms between the highest and the lowest
degrees, unless all even or all odd terms are missing.
The degree of P(s) and Q(s) may differ by zero, or one only.
The lowest degree in P(s) and Q(s) may differ in degree by at the most one.
15.12
NECESSARY CONDITIONS FOR TRANSFER FUNCTIONS
LO 6
The restrictions on pole and zero location in transfer functions with common factors in P(s) and Q(s) cancelled
are listed below.
1.
2.
3.
4.
5.
6.
7.
(a) The coefficients in the polynomials P(s) and Q(s) of N(s) 5 P(s)/Q(s) must be real.
(b) The coefficients in Q(s) must be positive, but some of the coefficients in P(s) may be negative.
Complex or imaginary poles and zeros must occur in conjugate pairs.
The real part of poles must be negative, or zero. If the real part is zero, then the pole must be simple.
The polynomial Q(s) may not have any missing terms between the highest and the lowest degree,
unless all even or all odd terms are missing.
The polynomial P(s) may have missing terms between the lowest and the highest degree.
The degree of P(s) may be as small as zero, independent of the degree of Q(s).
(a) For the voltage transfer ratio and the current transfer ratio, the maximum degree of P(s) must
equal the degree of Q(s).
(b) For transfer impedance and transfer admittance, the maximum degree of P(s) must equal the
degree of Q(s) plus one.
Frequently Asked Questions linked to LO 6
rrr 14-6.1
Explain the complete procedure for making a bode plot for different types of transfer functions.
rrr
rrr
s
R
Input
C
R
L
Fig. Q.3
s
rrr 14-6.5 Write the necessary conditions for driving-point function and transfer function.
rrr
690
15.13
Circuits and Networks
TIME-DOMAIN RESPONSE FROM POLE ZERO PLOT
For the given network function, a pole zero plot can be drawn which gives useful
LO 7
information regarding the critical frequencies. The time-domain response can also be
obtained from pole zero plot of a network function. Consider an array of poles shown in
Fig. 15.28.
In Fig. 15.28, s1 and s3 are complex conjugate poles, whereas s2 and s4 are
real poles. If the poles
are real, the quadratic function is
s 2 + 2dvn s + v2n for d > 1
where d is the damping ratio and vn is the
undamped natural frequency.
The roots of the equation are
s2 , s4 = -dvn ± vn d2 -1; d > 1
Fig. 15.28
For these poles, the time domain response is given by
i(t) 5 k2es2t 1 k4es4t
The response due to the pole s4 dies faster compared to that of s2 as
shown in Fig. 15.29.
s1 and s3 constitute complex conjugate poles. If the poles are
complex conjugate, then the quadratic function is
s2 1 2dvns 1 v2n for d , 1
The roots are s1, s1* = -dvn ± jvn 1- d2 for d < 1
Fig. 15.29
For these poles, the time domain response is given by
i(t ) = k1e-dvnt
(
) + k *e-dvnt - j(vn
1
+ j vn 1-d2 t
(
)
1-d2 t
)
= ke-dvnt sin vn 1- d2 t
From the above equation, we can conclude that the response for the conjugate poles is damped sinusoid.
Similarly, s3, s3* are also a complex conjugate pair.
Here, the response due to s3 dies down faster than that
due to s1 as shown in Fig. 15.30.
Consider a network having transfer admittance Y(s).
If the input voltage V(s) is applied to the network, the
corresponding current is given by
P ( s)
I ( s ) = V ( s ) Y ( s) =
Fig. 15.30
Q ( s)
This may be taken as
( s - sa )( s - sb ) …( s - sn )
I ( s) = H
( s - s1 )( s - s2 ) …( s - sm )
where H is the scale factor.
S-Domain Analysis
691
By taking the partial fractions, we get
k
k
k
I ( s) = 1 + 2 + … + m
s - s1 s - s2
s - sm
The time-domain response can be obtained by taking the inverse transform
 k
k 
k
i(t ) = L -1 1 + 2 +…+ m 
 s - s1 s - s2
s - sm 

Any of the above coefficients can be obtained by using the Heavisides method.
To find the coefficient kl,
 ( s - sa )( s - sb ) …( s - sn ) 
 ( s - sl )
k1 = H 
 ( s - s1 )( s - s2 ) …( s - sm ) 


s= sl
Here, sl, sm, sn are all complex numbers, the difference of (sl – sn) is also a complex number.
(sl – sn) 5 Mln e jfln
Hence k1 = H
M la M lb … M ln
× e j ( fla +flb +…+fln )-( fl1 +fl 2 +…+flm )
M l1 M l 2 … M lm
Similarly, all coefficients k1, k2, , km may be obtained, which constitute the magnitude and phase angle.
The residues may also be obtained by pole zero plot in the following way.
1. Obtain the pole zero plot for the given network function.
2. Measure the distances Mla, Mlb, , Mln of a given pole from each of the other zeros.
3. Measure the distances Ml1, Ml2, , Mlm of a given pole from each of the other poles.
4. Measure the angle fla, flb, ..., fln of the line joining that pole to each of the other zeros.
5. Measure the angle fl1, fl2, ..., flm of the line joining that pole to each of the other poles.
6. Substitute these values in required residue equation.
15.14
AMPLITUDE AND PHASE RESPONSE FROM POLE ZERO PLOT
LO 7
The steady-state response can be obtained from the pole zero plot, and it is given by
N( jv) 5 M(v)e jf(v)
where M(v) is the amplitude
f(v) is the phase
These amplitude and phase responses are useful in the design and analysis of network functions. For different
values of v, corresponding values of M(v) and f(v) can be obtained and these are plotted to get amplitude and
phase response of the given network.
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 7
rrr 15-7.1 For the given network function, draw the pole zero diagram and hence, obtain the time domain response. Verify
the result analytically.
I ( s) =
5s
( s + 3)( s + 2 s + 2)
2
692
Circuits and Networks
Frequently Asked Questions linked to LO 7
Vs
r
3s
(s + 1) (s + 4)
S
Is
rr
Is
RLC
rr
R
Is
rr
S
s+
L
Is
s+
s + s+
C
F
s
s+
s
rr
Hs
s
s
s
xt
s
e
t
YS
r
it
YS
1F
i(t)
4W
2H
2‡
Fig. Q.6
15.15
STABILITY CRITERION FOR AN ACTIVE NETWORK
Passive networks are said to be stable only when all the poles lie in the left half of
the s-plane. Active networks (containing controlled sources) are not always stable.
Consider transformed active network shown in Fig. 15.31.
By applying the Millman theorem, we get
V2 ( s) =
=
Fig. 15.31
V1 ( s) + kV2 ( s)
6+5 / s+ s
s [V1 ( s) + kV2 ( s)]
s2 + 6s + 5
V2(s) [s2 1 6s 1 5] – ksV2(s) 5 sV1(s)
V2(s) [s2 1 (6 – k)s 1 5] 5 sV1(s)
∴
LO 8
V2 ( s)
s
= 2
V1 ( s) s + (6 - k ) s + 5
From the above transformed equation, the poles are dependent upon the value of k.
S-Domain Analysis
693
The roots of the equation are
s=
-(6 - k ) ± (6 - k ) 2 - 4 ×5
2
For k 5 0, the poles are at –1, –5, which lie on the left half of the s-plane. As k increases, the poles move
towards each other and meet at a point
(6 - k ) 2 - 20 = 0 , when k 5 1.53 or 10.47. The root locus plot is
shown in Fig. 15.32.
The root locus is obtained from the characteristic
equation s2 1 (6 – k)s 1 5 5 0. As the value of
k increases beyond 1.53, the locus of the root is
a circle. The poles are located on the imaginary
axis at 6j2.24 for k 5 6. At –2.24, the poles are
coincident for k 5 1.53 while at 12.24, the poles
are coincident for k 5 10.47. When k . 10.47, the
poles again lie on the real axis but remain on the
right half of the s-plane, one pole moving towards
the origin and the other moving towards infinity.
From this, we can conclude, as long as k is less
than 6, the poles lie on the left half of the s-plane
and the system is said to be stable. For k 5 6, the
poles lie on the imaginary axis and the system is
Fig. 15.32
oscillatory in nature. For values of k greater than 6,
the poles lie on the right half of the s-plane. Then the system is said to be unstable.
15.16
ROUTH CRITERIA
LO 8
The locations of the poles gives us an idea about stability of the active network. Consider the denominator
polynomial
Q(s) 5 b0 sm 1 b1sm – 1 1 ... 1 bm
(15.78)
To get a stable system, all the roots must have negative real parts. There should not be any positive or zero
real parts. This condition is not sufficient.
Let us consider the polynomial
s3 1 4s2 1 15s 1 100 5 (s 1 5) (s2 – s 1 20)
In the above polynomial, though the coefficients are positive and real, the two roots have positive
real parts. From this, we conclude that the coefficients of Q(s) being positive and real is not a sufficient
condition to get a stable system. Therefore, we have to seek the condition for stability which is necessary
and sufficient.
Consider the polynomial Q(s) 5 0. After factorisation, we get
b0 (s – s1) (s – s2) ... (s – sm) 5 0
(15.79)
694
Circuits and Networks
On multiplication of these factors, we get
Q(s) 5 b0sm – b0(s1 1 s2 1 ... 1 sm)sm – 1
1 b0(s1s2 1 s2s3 1 ... ) sm – 2
1 b0 (–1)m (s1s2 ... sm) 5 0
(15.80)
Equating the coefficients of Eqs (15.78) and (15.80), we have
b1
= -( s1 + s2 + … + sm )
b0
(15.81)
5 - sum of the roots
b2
= 1( s1s2 + s2 s3 + …)
b0
(15.82)
5 sum of the products of the roots taken two at a time
b3
= -( s1s2 s3 + s2 s3 s4 + …)
b0
(15.83)
5 - sum of the products of the roots taken three at a time.
(-1) m
bm
= ( s1s2 s3 … sm ) 5 product of the roots
b0
(15.84)
If all the roots have negative real parts, then from the above equations, it is clear that all the coefficients
must have the same sign. This condition is not sufficient due to the fact that the zero value of a coefficient
involves cancellation, which requires some root to have positive real parts.
The Routh criterion for stability is discussed below. Consider a polynomial
Q(s) 5 b0sm 1 b1sm – 1 1 b2sm – 2 1 ... 1 bm
Taking first-row coefficients and second-row coefficients separately, we have
b0
b2
b4
...
b1
b3
b5
...
Now, we complete the Routh array as follows.
For m 5 5,
s5
b0
b2
b4
s
4
b1
b3
b5
s
3
c1
c2
s2
d1
d2
1
e1
0
f1
s
s
S-Domain Analysis
695
where c1, c2, d1, d2, el, f1 are determined by the algorithm given below.
b0
c1 =
c2 =
b2
b1
b3
b1
b0
b4
b1
b5
b1
=
b1b2 - b0 b3
b1
=
b1b4 - b0 b5
b1
b1 b3
c b -b c
c c
d1 = 1 2 = 1 3 1 2
c1
c1
b1 b5
b c -0
c 0
d2 = 1
= 51
c1
c1
e1 =
f1 =
c1
d1
c2
d2
d1
d1
e1
=
c2 d1 - c1d2
d1
=
d2 e1 - 0
e1
d2
0
e1
In order to find out the element in the kth row and jth column, it is required to know the four elements. These
elements in the row (k – 1) and the row (k – 2) just above the elements are in the column 1 of the array and the
(J 1 1) column of the array. The product of the elements joined by a line with positive slope has positive sign while
the product of elements joined with a line with negative slope has a negative sign. The difference of these products
is divided by the element of the column 1 and row (k – 1). The above process is repeated till m 1 1 rows are found
in the Routh array.
According to the Routh–Hurwitz theorem, the number of changes in the sign of the first column to the right
of the vertical line in an array moving from top to bottom is equal to the number of roots of Q(s) 5 0 with
positive real parts. To get a stable system, the roots must have negative real parts.
According to the Routh-Hurwitz criterion, the system is stable, if and only if, there are no changes in signs
of the first column of the array. This requirement is, both the necessary and sufficient condition for stability.
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 8
For the given denominator polynomial of a network function, verify the stability of the network
using Routh criteria.
Q(s) 5 s4 1 s3 1 2s2 1 2s 1 12
rrr 15-8.2 Apply the Routh criterion to the following equations and determine the number of roots (i) with
positive real parts, (ii) with zero real parts, and (iii) with negative real parts:
(a) 6s3 1 2s2 1 5s 1 2 5 0
(b) s6 1 5s5 1 13s4 1 21s3 1 20s2 1 16s 1 8 5 0
(c) s6 – s5 – 2s4 1 4s3 – 5s2 1 21s 1 30 5 0
rrr 15-8.1
696
Circuits and Networks
Frequently Asked Questions linked to LO 8
r
Ps
rr
s
s
s
s
r
Ps s
s
s
s
(a) With positive real roots
(b) With zero real parts
s
Additional Solved Problems
PROBLEM 15.1
For the two-port network shown in Fig. 15.33, determine the driving-point impedance z11(s), the transfer
impedance z21(s) and the voltage transfer ratio G21(s).
Fig. 15.33
Solution‰ The transformed circuit is shown in Fig. 15.34.
Fig. 15.34
From the above figure, by application of Kirchhoff’s laws, we get
V1 ( s) = (5 + 2 s) I1 ( s) - 2 sI 2 ( s)

2
0 = I 2 ( s)2 + 2 s +  - I1 ( s) 2 s

s
2
V2 ( s) = I 2 ( s)
s
(15.85)
(15.86)
(15.87)
697
S-Domain Analysis
From Eq. (15.86),
I 2 ( s) =
2s2
2s2 + 2s + 2
I1 ( s)
(15.88)
Substituting Eq. (15.88) in Eq. (15.85),
4 s3
V1 ( s) = (5 + 2 s) I1 ( s) - 2
I1 ( s)
2s + 2s + 2
The driving-point impedance,
V ( s) 7 s 2 + 7 s + 5
Z11 ( s) = 1
= 2
I1 ( s)
s + s +1
From Eqs (15.87) and (15.88),
4s
V2 ( s) = 2
I1 ( s)
2s + 2s + 2
The transfer impedance at the port 2 is
V ( s)
2s
Z 21 ( s) = 2
= 2
I1 ( s) s + s + 1
The voltage transfer ratio is
V ( s)
2s
G21 ( s) = 2
=
V1 ( s) 7 s 2 + 7 s + 5
PROBLEM 15.2
For the network shown in Fig. 15.35, determine the following transfer functions: (a) G21(s) (b) Z21(s).
Fig. 15.35
Solution‰ The transformed circuit is shown in Fig.
15.36.
The voltage across the port 2 is
V2 ( s) = V1 ( s)
2/s
2
2s
+ 2
s s +1
Fig. 15.36
698
Circuits and Networks
The voltage transfer ratio at the port 1 is
V ( s)
s2 +1
G12 ( s) = 1
= 2
V2 ( s) 2 s + 1
The voltage at the port 1 is
V1 ( s) = I ( s)
2
s
2 2
I 2 ( s) × 
 s  s
4 ( s2 + 1)
where V1 ( s) =
= I 2 ( s)
2
2
2s
s (6 s2 + 4)
+ 2
+
s s +1 s
The transfer impedance at the port 1 is
Z12 ( s) =
V1 ( s)
s2 +1
=
3

I 2 ( s)
s  s 2 + 1
2

PROBLEM 15.3
For the network shown in Fig. 15.37, determine transfer
impedance Z21(s) and Y21(s). Also find the transfer voltage
ratio G21(s) and the transfer current ratio a21(s).
Fig. 15.37
Solution‰ The transformed circuit is shown in Fig. 15.38.
The voltage at the port 1 is
V1 ( s) = I1 ( s)
5s
s +1
2
The voltage at the port 2 is
Fig. 15.38
The voltage transfer ratio at the port 2 is
V ( s)
G21 ( s) = 2
=1
V1 ( s)
The transfer impedance at the port 2 is
V ( s)
5s
Z 21 ( s) = 2
= 2
I1 ( s) s + 1
V2 ( s) = I1 ( s)
5s
s2 +1
S-Domain Analysis
699
The transfer admittance at the port 1 is
I ( s) s 2 + 1
Y12 ( s) = 1
=
V2 ( s)
5s
PROBLEM 15.4
For the given network function, draw the pole zero diagram and hence obtain the time-domain response.
Verify this result analytically.
I (s ) =
3s
(s + 1)(s + 3)
Solution‰ In the network function,
P ( s) = 3s
and Q(s) = ( s + 1)( s + 3)
By taking partial fractions, I(s) can be written as
A
B
I ( s) =
+
s +1 s + 3
Therefore, the time-domain response is
i(t) 5 Ae2t 1 Be23t
Here, the coefficients A and B are determined by using the pole zero plot as shown in Fig.15.39.
Consider the pole at –1
The distance between zero to pole at –1 is
M01 5 1
The angle between the line joining the pole at –1 to the zero is
f01 5 180°
Similarly, the distance between the pole at –3 to the pole to –1 is
M31 5 2
The angle between the line joining the pole at –1 to the pole at –3
is
Fig. 15.39
f31 5 0°
H53
Hence,
A=H
M 01 e jf01
M 31 e
Similarly, B = H
where
jf31
M 03 e jf03
M13 e
jf13
=
-3
2
=
9
2
M 03 = 3 ; f03 = 180°
M13 = 2 ; f13 = 180°
700
Circuits and Networks
Substituting these values, we get
 -3

9
i(t ) =  e-t + e-3t  A
 2

2
Analytically,
I ( s) =
3s
( s + 1)( s + 3)
By taking partial fractions,
I ( s) =
A
B
+
s +1 s + 3
-3
2
9
B = I ( s)( s + 3) s=-3 =
2
By substituting these values in the above equation ,
9
-3
I ( s) =
+
(
)
(
2 s + 1 2 s + 3)
A = I ( s)( s + 1) s=-1 =
Taking inverse transform, we get
i (t ) =
-3 -t 9 -3t
e + e
2
2
PROBLEM 15.5
For the given denominator polynomial of a network function, verify the stability of the network using Routh
criteria.
Q(s) 5 s5 1 3s4 1 4s4 1 5s2 1 6s 1 1
Solution‰ The Routh array for this polynomial is given below.
s5
1
4
6
s
4
3
5
1
s
3
2.33 5.67
s
2
-2.3
s1
1
6.68
Since there is –ve sign in the first column, the system is unstable.
PROBLEM 15.6
Draw the pole zero diagram for the given network function and hence, obtain n(t).
V (s ) =
4(s + 2)s
(s + 1)(s + 3)
S-Domain Analysis
701
Solution‰ In the network function,
p(s) 5 4s(s 1 2)
and Q(s) 5 (s 1 1) (s 1 3) 5 0
By taking partial fractions, we have
k
k
V ( s) = 1 + 2
s +1 s + 3
The time-domain response can be obtained by taking the inverse transform
v(t) 5 k1 e–t 1 k2 e–3t
Here, the coefficients k1 and k2 may be determined
by using the pole zero plot as shown in Fig. 15.40.
To determine k1, we have to find out the distances
and phase angles from other zeros and poles to that
particular pole.
M M e j ( f01 + f21 )
Hence, k1 = H 01 21 j ( f )
M 31e 31
Fig. 15.40
where M01 and M21 are the distances between the
zeros at 0 and –2 to the pole at –1, f01, f21 are the
phase angle between the corresponding zeros to the pole.
Similarly, M31 and f31 are the distance and phase angle, respectively, from the pole at –3 to the pole at –1.
∴ M01 5 1; f01 5 180°
M21 5 1; f21 5 0
M31 5 2; f31 5 0°
∴
k1 = 4 ×
k1 5 –2
1×1 j (180°)
e
2
Similarly,
k2 = H
M 03 M 23 + j ( f03 + f23 - f13 )
e
M13
where M03 5 3, f03 5 180°
M23 5 1, f23 5 180°
M13 5 2, f13 5 180°
4 ×3×1 j (180 + 180 - 180 )
e
2
k2 5 –6
∴ k2 =
Substituting the values, we get
(t) 5 (–2e–t – 6e–3t) V
702
Circuits and Networks
PROBLEM 15.7
For the given network function, draw the pole zero diagram and hence, obtain the time-domain response i(t).
I (s ) =
5s
(s + 1)(s + 4s + 8)
2
Solution‰ In the network function,
P (s) 5 5s
Q(s) 5 (s 1 1) (s2 1 4s 1 8) 5 0
By taking the partial fraction expansion of I(s), we get
k3
k
k2
I ( s) = 1 +
+
s + 1 ( s + 2 + j 2) ( s + 2 - j 2)
(15.89)
The time-domain response can be obtained by taking the inverse transform as under,
i(t) 5 k1e–t 1 k2 e–(2 1 j2)t 1 k2 e–(2 – j2)t
(15.90)
To find the value of k1, we have to find out the
distances and phase angles from other zeros and poles
to that particular pole as shown in Fig. 15.41.
Hence, k1 =
M p11 M p 21e
Fig. 15.41
M01 5 1; f01 5 180°
∴
Mp11 5
5 ; fp11 5 –63.44°
Mp21 5
5 ; fp21 5 63.44°
k1 =
5×1e j180°
5 × 5e j (-63.44°+63.44°)
k1 5 –1
Similarly, k2 =
HM 0 p1 e
M1 p1 M p2 p1 e
jf 0 p 1
j ( f1 p1 +f p2 p 1 )
M0p 5 8 ; f0p 5 135°
1
1
M1p 5 5 ; f1p 5 116.56°
1
1
Mp p 5 4; fp p 5 90°
1 2
Hence, k2 =
2
5× 8
5 ×4
H M 01e j ( f01 )
1
e j (135°-116.56°-90°)
j f p11 +f p 21 
S-Domain Analysis
703
5 1.58 e–j(71.56°)
k2* =
=
H M 0 p2 e
M1 p2 M p1 p2 e
j f0 p2
j ( f1 p2 +f p1 p2 )
5× 8 e- j (135°)
5 × 4e j (-116.56°-90°)
= 1.58e j 71.56°
If we substitute the values in Eq. (15.90), we get
i(t) 5 [–1e–t 1 1.58 e–j(71.56°) e–(2 1 j2)t 1 1.58e j(71.56°) e–(2 – j2)t] A
PROBLEM 15.8
For the given denominator polynomial of a network function, verify the stability of the network by using the
Routh criterion.
Q(s) 5 s3 1 2s2 1 8s 1 10
Solution‰ The Routh array for this polynomial is given below.
s3
s2
1
2
8
10
s1 3
s0 10
There is no change in sign in the first column of the array. Hence, there are no roots with positive real parts.
Therefore, the network is stable.
PROBLEM 15.9
For the given denominator polynomial of a network function, verify the stability of the network using the
Routh criterion.
Q(s) 5 s3 1 s2 1 3s 1 8
Solution‰ The Routh array for this polynomial is given below.
s3
1 3
s2
1
1
-5
s
8
s +8
There are two changes in sign of the first column, one from 1 to –5 and the other from –5 to 18. Therefore, the
two roots have positive real parts. Hence, the network is not stable.
0
PROBLEM 15.10
For the given denominator polynomial of a network function, determine the value of k for which the network
to stable.
Q(s) 5 s3 1 2s2 1 4s 1 k
704
Circuits and Networks
Solution‰ The Routh array for the given polynomial is given below.
s3
1
4
s2
2
8- k
2
k
k
s1
s0
When k , 8, all the terms in the first column are positive. Therefore, there is no sign change in the first
column. Hence, the network is stable. When k . 8, the 8 – k/2 is negative. Therefore, there are two sign
changes in the first column. There are two roots which have positive real parts. Hence, the network is unstable.
When k 5 8, the Routh array becomes
s3
1
4
2
2
8
1
a
s
s
0
s 8
The element in the first column and third row is zero. But we can take it as a small number. In this case,
there are no changes in the sign of the first column. Hence, the network is stable.
PROBLEM 15.11
Apply the Routh criterion to the given polynomial and determine the number of roots (a) with positive real
parts, (b) with zero real parts, and (c) with negative real parts.
Q(s) 5 s4 1 4s3 1 8s2 1 12s 1 15
Solution‰ The Routh array for the polynomial is
s4
1
8
3
4
12
2
5
15
1
0
0
0
?
?
s
s
s
s
15
In this case, all the elements in the fourth row have become zero and the array cannot be completed.
The given equation is reduced by taking the new polynomial from the third row
5s2 1 15 5 0
5(s2 1 3) 5 0
Hence, the other polynomial
Q2 ( s) =
s 4 + 4 s3 + 8s 2 + 12 s + 15
5( s 2 + 3)
The equation reduces to the following polynomial.
(s2 1 3) (s2 1 4s 1 5) 5 0
705
S-Domain Analysis
The roots of the equation s2 1 3 5 0 are s = ± j 3
These two roots have zero real parts.
Again, forming the Routh array for the polynomial,
s 2 + 4s + 5 = 0
s2
1
5
1
4
0
0
5
s
s
There are no changes in the sign of the first column. Hence, all the roots have negative real parts. Therefore,
out of the four roots, two roots have negative real parts and two roots have zero real parts.
ANSWERS TO PRACTICE PROBLEMS
15-4.1
Z ( s) =
2 s 3 + 3s 2 + 2 s + 1
Y ( s) =
15-4.2
15-4.3
s3 + 2 s 2 + 3s + 2
s3 + 2 s 2 + 3s + 2
2 s + 3s + 2 s + 1
s2 + 2s + 1
G21 ( s) = 2
s + 3s + 1
zd ( s ) =
15-5.1
3
2
10 s 2 + 27 s + 30
5s + 4
G 21 ( S ) =
2s
7s + 7s + 5
2
15-5.2 The zeros are lying at s 5 –0.5 and s 5 –0.04
-2 t
+ 12 e-3t
15-5.3 v (t ) = -8 e
15-5.4
1
2
1- e-t / 3 + e-( t -3)/ 5
3
15
15-7.1
The poles are lying at s 5 –0.98 and s 5 –0.02
15-8.2
i (t) 5 4.5e23t – 1.5e2t
Objective-Type Questions
rrr 15.1
rrr 15.2
The driving-point impedance is defined as
(a) the ratio of transform voltage to transform current at the same port
(b) the ratio of transform voltage at one port to the transform current at the other port
(c) both (a) and (b)
(d) none of the above
The transfer impedance is defined as
(a) the ratio of transform voltage to transform current at the same port
(b) the ratio of transform voltage at one port to the current transform at the other port
(c) both (a) and (b)
(d) none of the above
rrr 15.3
The function is said to be having simple poles and zeros only if
(a) the poles are not repeated
(b) the zeros are not repeated
(c) both poles and zeros are not repeated
(d) none of the above
706
rrr 15.4
Circuits and Networks
The necessary condition for a driving-point function is
(a) the real part of all poles and zeros must not be zero or negative
(b) the polynomials P(s) and Q(s) may not have any missing terms between the highest and lowest degree
unless all even or all odd terms are missing
(c) the degree of P(s) and Q(s) may differ by more than one
(d) the lowest degree in P(s) and Q(s) may differ in degree by more than two
rrr 15.5
The necessary condition for the transfer functions is that
(a) the coefficients in the polynomials P(s) and Q(s) must be real
(b) coefficients in Q(s) may be negative
(c) complex or imaginary poles and zeros may not conjugate
(d) if the real part of pole is zero, then that pole must be multiple
rrr 15.6
The system is said to be stable, if and only if
(a) all the poles lie on the right half of the s-plane
(b) some poles lie on the right half of the s-plane
(c) all the poles does not lie on the right half of the s-plane
(d) none of the above
For interactive quiz with answers,
scan the QR code given here
OR
visit
http://qrcode.flipick.com/index.php/273
16
LEARNING OBJECTIVES
p
16.1
TWO-PORT NETWORK
Generally, any network may be represented schematically by a rectangular box.
LO 1
A network may be used for representing either source or load, or for a variety of
purposes. A pair of terminals at which a signal may enter or leave a network is
called a port. A port is defined as any pair of terminals into which energy is supplied, or from which energy
is withdrawn, or where the network variables may be measured. One such network having only one pair of
terminals (1-19) is shown in Fig. 16.1 (a).
(a)
(b)
Fig. 16.1
708
Circuits and Networks
A two-port network is simply a network inside a black box, and the network has only two pairs of accessible
terminals; usually one pair represents the input and the other represents the output. Such a building block is
very common in electronic systems, communication systems, transmission, and distribution systems. Figure
16.1(b) shows a two-port network, or a two terminal pair network, in which the four terminals have been paired
into ports 1-19 and 2-29. The terminals 1-19 together constitute a port. Similarly, the terminals 2-29 constitute
another port. Two ports containing no sources in their branches are called passive ports; among them are power
transmission lines and transformers. Two ports containing sources in their branches are called active ports.
A voltage and current assigned to each of the two ports. The voltage and current at the input terminals are V1 and
I1; whereas V2 and I2 are specified at the output port. It is also assumed that the currents I1 and I2 are entering
into the network at the upper terminals 1 and 2, respectively. The variables of the two-port network are V1, V2,
and I1, I2. Two of these are dependent variables, the other two are independent variables. The number of possible
combinations generated by the four variables, taken two at a time, is six. Thus, there are six possible sets of
equations describing a two-port network.
16.2
OPEN-CIRCUIT IMPEDANCE (Z) PARAMETERS
A general linear two-port network defined in Section 16.1 which does not LO 2
contain any independent sources is shown in Fig. 16.2.
The Z-parameters of a two‑port for the positive directions of voltages and
currents may be defined by expressing the port voltages V1 and V2 in terms of
the currents I1 and I2. Here,
V1 and V2 are dependent
variables, and I1, I2 are
independent variables. The voltage at port 1‑19 is the
response produced by the two currents I1 and I2. Thus,
V1 5 Z11 I1 1 Z12 I2
Fig. 16.2
Similarly,
V2 5 Z21 I1 1 Z22 I2
(16.1)
(16.2)
Z11, Z12, Z21, and Z22 are the network functions, and are called impedance (Z ) parameters, and are
defined by Eqs (16.1) and (16.2). These parameters can be represented by matrices.
We may write the matrix equation as [V ] 5 [Z ] [I ]
V 
where V is the column matrix 5  1 
V2 
 
 Z11
Z is the square matrix 5  Z
 21
Z12 

Z 22 
I 
and we may write |I | in the column matrix 5  1 
I2 
 
Thus,
V1   Z11
 =
V2   Z 21
  
Z12   I1 
 
Z 22   I 2 
Two-Port Networks
709
The individual Z-parameters for a given network can be defined by setting each of the port currents
equal to zero. Suppose the port 2-29 is left open-circuited, then I2 5 0.
Thus,
Z11 =
V1
I1
I 2 =0
where Z11 is the driving-point impedance at the port 1-19 with the port 2-29 open circuited. It is called
the open‑circuit input impedance.
Similarly,
Z21 =
V2
I1
I 2 =0
where Z21 is the transfer impedance at the port 1-19 with the port 2-29 open-circuited. It is also called the
open‑circuit forward transfer impedance. Suppose the port 1-19 is left open circuited, then I1 5 0.
Thus, Z12 =
V1
I2
I1 =0
where Z12 is the transfer impedance at the port 2-29, with the port 1-19 open- circuited. It is also called
the open‑circuit reverse transfer impedance.
Z 22 =
V2
I2
I1 =0
where Z22 is the open-circuit driving-point impedance at the port 2-29 with the port 1-19 open circuited. It
is also called the open‑circuit output impedance. The equivalent circuit of the two-port networks governed
by Eqs (16.1) and (16.2), i.e. open-circuit impedance parameters is shown in Fig. 16.3.
Fig. 16.3
If the network under study is reciprocal or bilateral, then in accordance with the reciprocity principle,
V2
I1
or
=
I 2 =0
V1
I2
I1 =0
Z21 5 Z12
It is observed that all the parameters have the dimensions of impedance. Moreover, individual parameters
are specified only when the current in one of the ports is zero. This corresponds to one of the ports being
open-circuited from which the Z-parameters also derive the name open‑circuit impedance parameters.
710
Circuits and Networks
EXAMPLE 16.1
Find the Z‑parameters for the circuit shown in Fig. 16.4.
Fig. 16.4
Solution The circuit in the problem is a T-network. From Eqs (16.1) and (16.2) we have
V1 5 Z11 I1 1 Z12 I2
V2 5 Z21 I1 1 Z22 I2
When the port b-b9 is open-circuited, Z11 =
V1
I1
where V1 5 I1(Za 1 Zb)
∴ Z11 5 (Za 1 Zb)
Z 21 =
V2
I1
I 2 =0
where V2 5 I1 Zb
∴ Z21 5 Zb
When the port a-a9 is open-circuited, I1 5 0
Z 22 =
V2
I2
I1 =0
where V2 5 I2(Zb 1 Zc)
∴ Z22 5 (Zb 1 Zc)
Z12 =
V1
I2
I1 =0
where V1 5 I2 Zb
∴ Z12 5 Zb
It can be observed that Z12 5 Z21, so the network is a bilateral network which satisfies the principle of
reciprocity.
Two-Port Networks
16.3
SHORT-CIRCUIT ADMITTANCE (Y) PARAMETERS
711
LO 2
A general two-port network which is considered in Section 16.2 is shown in Fig. 16.5.
Fig. 16.5
The Y-parameters of a two‑port network for the positive directions of voltages and currents may be defined
by expressing the port currents I1 and I2 in terms of the voltages V1 and V2. Here, I1, I2 are dependent variables
and V1 and V2 are independent variables. I1 may be considered to be the superposition of two components,
one caused by V1 and the other by V2.
Thus,
I1 5 Y11 V1 1 Y12 V2
(16.3)
Similarly, I2 5 Y21 V1 1 Y22 V2
(16.4)
Y11, Y12, Y21, and Y22 are the network functions and are also called the admittance (Y ) parameters. They are
defined by Eqs (16.3) and (16.4). These parameters can be represented by matrices as follows:
[I ] 5 [Y ] [V ]
where
I 
Y
Y 
I =  1  ; Y =  11 12 
I2 
Y21 Y22 
 


and
V 
V =  1
V2 
 
 I1  Y11 Y12  V1 
  
 
Thus,  I  = Y
 V2 
Y
21
22
2
  
 
The individual Y-parameters for a given network can be defined by setting each port voltage to zero. If we
let V2 be zero by short-circuiting the port 2-29, then
Y11 =
I1
V1 V =0
2
Y11 is the driving-point admittance at the port 1-19, with the port 2-29 short- circuited. It is also called the
short‑circuit input admittance.
Y21 =
I2
V1
V2 =0
712
Circuits and Networks
Y21 is the transfer admittance at the port 1-19 with the port 2-29 short-circuited. It is also called short‑circuited
forward transfer admittance. If we let V1 be zero by short- circuiting the port 1-19, then
Y12 =
I1
V2
V1 =0
Y12 is the transfer admittance at the port 2-29 with the port 1-19 short-circuited. It is also called the short‑circuit
reverse transfer admittance.
Y22 =
I2
V2
V1 =0
Y22 is the short-circuit driving-point admittance at the port 2-29 with the port 1-19 short circuited. It is also
called the short‑circuit output admittance. The equivalent circuit of the network governed by Eqs (16.3) and
(16.4) is shown in Fig . 16.6.
Fig. 16.6
If the network under study is reciprocal, or bilateral, then
I1
V2
or
=
V1 =0
I2
V1
V2 =0
Y12 5 Y21
It is observed that all the parameters have the dimensions of admittance which are obtained by short-circuiting
either the output or the input port from which the parameters also derive their name, i.e. the short‑circuit
admittance parameters.
EXAMPLE 16.2
Find the Y‑parameters for the network shown in Fig. 16.7.
Fig. 16.7
Two-Port Networks
Solution
Y11 =
I1
V1 V =0
2
When b-b9 is short-circuited, V2 5 0 and the network looks as shown in Fig. 16.8 (a).
Fig. 16.8 (a)
V1 5 I1 Zeq
Zeq 5 2 V
∴ V1 5 I1 2
Y11 =
I1 1
= 
V1 2
Y21 =
I2
V2
V2 =0
2 I
With the port b-b9 short-circuited, −I 2 = I1 × = 1
4
2
V1
∴ −I 2 =
4
I2
1
Y21 =
=− 
V1 V =0
4
2
Similarly, when the port a-a9 is short circuited, V1 5 0 and the network looks as shown in Fig. 16.8 (b).
Fig. 16.8 (b)
I
Y22 = 2
V2
V1 =0
V2 5 I2 Zeq
where Zeq is the equivalent impedance as viewed from b-b9.
713
714
Circuits and Networks
Zeq =
8
V
5
V2 = I 2 ×
Y22 =
Y12 =
8
5
I2
V2
V1 =0
I1
V2
V1 =0
=
5

8
With a-a9 short-circuited, −I1 =
2
I2
5
5V2
8
V
2 5
−I1 = × V2 = 2
5 8
4
I1
1
=− 
∴ Y12 =
V2
4
Since
I2 =
The describing equations in terms of the admittance parameters are
I1 5 0.5 V1 – 0.25 V2
I2 5 – 0.25 V1 1 0.625 V2
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 2*
rrr 16-2.1
rrr 16-2.2
Find the Z-parameters of the network shown in Fig. Q.1.
Determine the impedance parameters for the T‑network shown in Fig. Q.2 and draw the
Z‑parameter equivalent circuit.
j
Fig. Q.1
*Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
Fig. Q.2
715
Two-Port Networks
rrr 16-2.3
Determine the input and output impedances for the Z-parameter equivalent circuit shown in Fig.
Q.3.
rrr 16-2.4
The Z-parameters of a two-port network shown in Fig. Q.4 are Z11 5 5 V; Z12 5 4 V; Z22 5 10
V; Z21 5 5 V. If the source voltage is 25 V, determine I1, V2 I2, and the driving-point impedance
at the input port.
Fig. Q.3
Fig. Q.4
rrr16-2.5
For the network shown in Fig.
Q.5, determine all four opencircuit impedance parameters.
Find the inverse transmission
parameters for the network in
Fig. Q.6.
rrr 16-2.7 Determine
the
admittance
parameters for the p-network
shown in Fig. Q.7 and draw the
Y-parameter equivalent circuit.
rrr 16-2.6
Fig. Q.6
Fig. Q.5
Fig. Q.7
716
Circuits and Networks
rrr 16-2.8
For the network shown in Fig. Q.8, determine y12 and y21.
Fig. Q.8
rrr16-2.9 For the network shown in Fig. Q.9, determine Y-parameters.
rrr 16-2.10 Using PSpice, obtain g-parameters of the two-port network shown
Fig. Q.9
rrr 16-2.11 Find
in Fig. Q.10.
Fig. Q.10
Y-parameters of the network shown in Fig. Q.11.
I2
I1
V2
V1
Fig. Q.11
rrr 16-2.12 Find
Z‑ and Y-parameters of the given -network. (Fig. Q. 12)
rrr 16-2.13 Find the Y-parameters of the two-port network shown in Fig. Q.13.
1
1
Fig. Q.12
Fig. Q.13
717
Two-Port Networks
Frequently Asked Questions linked to LO 2*
Z
rrr
s
Z
s
Y
rrr
2F
1W
1W
V1
V2
2F
Fig. Q.1
rrr 16-2.3
rrr 16-2.4
rrr 16-2.5
rrr 16-2.6
rrr 16-2.7
rrr 16-2.8
Fig. Q.2
1W
1W
Find the Z-parameters for the network shown
in Fig. Q.3.
[BPUT 2007]
I
I
Why are Z-parameters known as open-circuit
2W
parameters?
[BPTU 2008]
V
A two-port device is defined by the following V
pair of equations: 2V1 + V2 and i2 = V1 + V2.
Write its impedance parameters Z11, Z12, Z21,
Z22,.
[BPTU 2008]
Fig. Q.3
Explain reciprocal.
[GTU Dec. 2010]
Find the Z-parameters for the network shown in Fig. Q.7.
[GTU Dec. 2010]
Obtain Z-parameters and transmission parameters of the network shown in Fig. Q.8.
[JNTU Nov. 2012]
2
1
2
1
I1 10 W
V1
20 W I
2
5W
0,5V2
I1
V2
2W
+
–
3I1
4W
Fig. Q.7
I2
5W
V2
Fig. Q.8
rrr 16-2.9
[MU 2014]
[PTU 2011-12]
[PU 2010]
[RTU Feb. 2011]
Obtain transmission parameters in terms of Z-parameters.
rrr 16-2.10 Determine the expression for Z-parameters of lattice networks.
rrr 16-2.11 Find the Z-parameters for the network shown in Fig. Q.11.
rrr 16-2.12 Find the Z-parameters for the network shown in Fig. Q.12.
2W
1
I1
1W
1W
2W
V1
1W
I1
I2
2W
2W
V1
V2
+
Vx
–
2W
1W
+
–
1.5 Vx
1'
Fig. Q.11
2
I2
V2
2'
Fig. Q.12
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
718
Circuits and Networks
rrr 16-2.13
What is a reciprocal network? Derive the condition for reciprocity in terms of Z-parameters.
[RGTU Dec. 2013]
rrr 16-2.14 Find the Z-parameters for the circuit shown in the Fig. Q.14.
[RGTU Dec. 2013]
rrr 16-2.15 Find the Z-parameters for the network given in Fig. Q.15.
[RTU Feb. 2011]
I2
Za
Zc
1W
I1
2W
b
a
4W
V1
V2
2W
Zb
1
1
a
b
Fig. Q.14
rrr 16-2.16
rrr 16-2.17
Fig. Q.15
[BPUT 2007]
[BPUT 2008]
Find Y11(s) of the circuit shown in Fig. Q.16
Determine the Y-parameters of the given network shown in Fig. Q.17
1F I2
I1
R
2 ohms
1
C1
I1(s)
V1(s)
V2
V1
V2(s)
1H
Fig. Q.17
Fig. Q.16
Determine the voltage across the capacitor in the RLC circuit as shown in
Fig. Q.18, if R = 400 ohms using Laplace transform. [GTU May 2011]
rrr 16-2.19 Explain the short-circuit admittance and the open-circuit impedance
parameters for a two-port network.
[GTU May 2011]
rrr 16-2.20 Find the open-circuit impedance parameters of the circuit shown in Fig.
Q.20. Also find the Y-parameters.
[JNTU Nov. 2012]
rrr 16-2.21 Determine Y-parameters of the network shown in Fig. Q.21.
I1
V1
1W
2W
4W
2V1
Fig. Q.20
I1
I2
3 W V2
V1
10 mH
R
rrr 16-2.18
1 micro-F
+ u(t)
–
Fig. Q.18
[JNTU Nov. 2012]
I2
3W
2W
4W
3I2
V2
Fig. Q.21
rrr 16-2.22
Find the Y and Z parameters of the network in Fig. Q.22
[PTU 2009-10]
rrr 16-2.23 Derive the condition of reciprocity and symmetry for
y-parameters.
[PU 2010]
rrr 16-2.24 Obtain the reciprocity and symmetry conditions for
Z-and Y-parameters.
2W
V1
V2
2W
3W
V2
Fig. Q.22
[PU 2012]
719
Two-Port Networks
rrr 16-2.25
The network shown in Fig. Q.25 contains a current-controlled current source. For this network
find the Y-parameters.
[RGTU Dec. 2012]
rrr 16-2.26 Find the Y-parameters for the network of Fig. Q.26.
[RTU Feb. 2011]
I1
V1
1W
1W
I2
2W
2W
V1
3I1 V2
4W
2W
Fig. Q.25
16.4
2W
Fig. Q.26
TRANSMISSION (ABCD) PARAMETERS
Transmission parameters, or ABCD parameters, are widely used in transmissionLO 3
line theory and cascade networks. In describing the transmission parameters,
the input variables V1 and I1 at the port 1-19, usually called the sending end,
are expressed in terms of the output variables V2 and I2 at the port 2-29, called
the receiving end. The transmission parameters provide a direct relationship
between input and output. Transmission parameters are also called general circuit parameters, or chain
parameters. They are defined by
V1 5 AV2 – BI2
(16.5)
I1 5 CV2 – DI2
(16.6)
The negative sign is used with I2, and not for the parameters B and D. Both the port currents I1 and –
I2 are directed to the right, i.e. with a negative sign in Eqs (16.5) and (16.6), the current at the port 2-29
which leaves the port is designated as positive. The parameters A, B, C and D are called the transmission
parameters. In the matrix form, Eqs (16.5) and (16.6) are expressed as
V1   A B   V2 
 =


 I1  C D  −I 2 

  

 A B
 is called the transmission matrix.
The matrix 

C D 
For a given network, these parameters can be determined
as follows. With the port 2-29 open, i.e. I2 5 0; applying a
voltage V1 at the port 1-19, using Eq. (16.5), we have
V1
V2
I 2 =0
1 V2
=
A V1
I 2 =0
A=
Fig. 16.9
and C =
= g21 I
I1
V2
I 2 =0
2 =0
1/A is called the open-circuit voltage gain, a dimensionless parameter. And
1 V2
=
C
I1
= Z 21 , which is the
I 2 =0
open-circuit transfer impedance. With the port 2-29 short circuited, i.e. with V2 5 0, applying the voltage V1 at
720
Circuits and Networks
the port 1-19, from Eq. (16.6), we have
V
I
−B = 1
and − D = 1
I 2 V =0
I2
2
−
−
1 I2
=
B V1
V2 =0
I
1
= 2
D I1
V2 =0
V2 =0
= Y21 , which is the short-circuit transfer admittance
=  21 V =0 , which is the short-circuit current gain, a dimensionless parameter.
2
16.4.1 Cascade Connection
The main use of the transmission matrix is in dealing with a cascade connection of two-port networks as
shown in Fig. 16.10.
Fig. 16.10
Let us consider two two-port networks Nx and Ny connected in cascade with port voltages and currents as
indicated in Fig. 16.10. The matrix representation of ABCD parameters for the network X is as under.
V1   Ax
 =
 I1  C x
  
Bx   V2 x 


Dx  −I 2 x 
And for the network Y, the matrix representation is
V1 y   Ay
 =
 I  C
 1 y   y
B y   V2 y 


D y  −I 2 y 


It can also be observed that at 2-29,
V2x 5 V1y and I2x 5 – I1y.
Combining the results, we have
V1   Ax Bx   Ay B y   V2 


 =


 I1  C x Dx  C

D
y  −I1 
  
  y
V1   A B   V2 
 =


 I1  C D  −I 2 

  

 A B
 is the transmission-parameters matrix for the overall network.
where 


C D 
Thus, the transmission matrix of a cascade of a two-port networks is the product of transmission matrices
Two-Port Networks
721
of the individual two-port networks. This property is used in the design of telephone systems, microwave
networks, radars, etc.
EXAMPLE 16.3
Find the transmission or general circuit parameters for the circuit shown in Fig. 16.11.
Fig. 16.11
Solution From Eqs (16.5) and (16.6) in Section 16.4, we have
V1 5 AV2 – BI2
I1 5 CV2 – DI2
When b-b9 is open, I 2 = 0; A =
V1
V2
I 2 =0
where V1 5 6I1 and V2 5 5I1
∴
A=
6
5
C=
I1
V2
=
I 2 =0
1

5
When b-b9 is short-circuited; V2 5 0 (see Fig. 16.12)
B=
−V1
−I
;D= 1
I 2 V =0
I 2 V =0
2
2
Fig. 16.12
722
Circuits and Networks
−I 2 =
5
V1
17
∴
B=
17
V
5
Similarly,
I1 =
7
5
V1 and − I 2 =
V1
17
17
∴
D=
7
5
In the circuit,
16.5
INVERSE TRANSMISSION (A9 B9 C9 D9) PARAMETERS
LO 3
In the preceding section, the input port voltage and current are expressed in terms of output port voltage
and current to describe the transmission parameters. While defining the transmission parameters, it is
customary to designate the input port as the sending end and the output port as the receiving end. The
voltage and current at the receiving end can also be expressed in terms of the sending end voltage and
current. If the voltage and current at the port 2-29 are expressed in terms of voltage and current at the port
1-19, we may write the following equations.
V2 5 A9V1 – B9I1
(16.7)
I2 5 C9V1 – D9I1
(16.8)
The coefficients A9, B9, C9, and D9 in the above equations are called inverse transmission parameters.
Because of the similarities of Eqs (16.7) and (16.8) with Eqs (16.5) and (16.6) in Section 16.4, the A9, B9,
C9, D9 parameters have properties similar to ABCD parameters. Thus, when the port 1-19 is open, I1 5 0.
A′ =
V2
V1
; C′ =
I1 =0
I2
V1
I1 =0
If the port 1-19 is short-circuited, V1 5 0
B′ =
Fig. 16.13
−V2
I1
; D=
V1 =0
−I 2
I1
V1 =0
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 3
rrr 16-3.1
Find the transmission parameters for the R–C network shown in Fig. Q.1.
Fig. Q.1
723
Two-Port Networks
rrr 16-3.2
Calculate the overall transmission parameters for the cascaded network shown in Fig. Q.2.
Fig. Q.2
rrr 16-3.3
rrr 16-3.4
Determine the impedance parameters and the transmission parameters for the network in Fig. Q.3.
Using PSpice, find transmission parameters of the network shown in Fig. Q.4.
Fig. Q.3
Fig. Q.4
Frequently Asked Questions linked to LO 3
rrr 16-3.1
ABCD-parameters are also known as transmission parameters and they are derived from the basic
two-port network parameters. Show that, for reciprocal linear time invariant two-port network,
AD‑BC = 1.
[GTU Dec. 2010]
rrr 16-3.2 Find ABCD-parameters for the two-port network shown in Fig. Q.2. Also derive Y-parameters
from the ABCD-parameters.
[GTU Dec. 2012]
rrr 16-3.3 For the network shown in Fig. Q.3, find ABCD-parameters.
[JNTU Nov. 2012]
1
V1
2W
2W
W
1W
3W
2
I
2 W V2
2
1W
5W
2W
1'
Fig. Q.2
2'
I
1
2
1
Fig. Q.3
rrr 16-3.4
Find transmission parameters for the network shown in Fig.
Q.4.
[PU 2010]
rrr 16-3.5
Define ABCD parameters for a two-port network.
[RGTU June 2014]
I1
V1
1W
2W
5W
Fig. Q.4
I2
V2
724
Circuits and Networks
rrr 16-3-6
16.6
Find the ABCD-parameters of the network shown
in Fig. Q.6. Also find the image parameters for the
network.
[RTU Feb. 2011]
2W
V1
3W
5W
2W
Fig. Q.6
HYBRID (H) PARAMETERS
Hybrid parameters, or h-parameters find extensive use in transistor circuits. They
LO 4
are well suited to transistor circuits as these parameters can be most conveniently
measured. The hybrid matrices describe a two-port network, when the voltage of one
port and the current of other port are taken as the independent variables. Consider the
network in Fig. 16.14.
If the voltage at the port 1-19 and current at the port 2-29 are taken as dependent variables, we can express
them in terms of I1 and V2.
V1 5 h11 I1 1 h12 V2
(16.9)
I2 5 h21 I1 1 h22 V2
(16.10)
The coefficients in the above equations are called
hybrid parameters. In matrix notation,
V1   h11
 =
 I 2   h21
  
Fig. 16.14
h12   I1 
 
h22  V2 
From Eqs (16.9) and (16.10), the individual
h-parameters may be defined by letting I1 5 0 and V2 5 0.
When V2 5 0, the port 2-29 is short-circuited.
Then
h11 =
V1
= Short-circuit input impedance =
I1 V =0
2
h21 =
I2
I1
V2 =0
 1 
 
Y11 
Y 
= Short-circuit forward current gain =  21 
 Y 
11
Similarly, by letting port 1-19 open, I1 5 0
h12 =
h22 =
V1
V2
I2
V2
I1 =0
Z 
Open-circuit reverse voltage gain =  12 
 Z 
22
I1 =0
 1 
Open-circuit output admittance = 

 Z 22 
Since the h-parameters represent dimensionally an impedance, an admittance, a voltage gain, and a
current gain, these are called hybrid parameters. An equivalent circuit of a two-port network in terms of
725
Two-Port Networks
hybrid parameters is shown in Fig. 16.15.
Fig. 16.15
EXAMPLE 16.4
Find the h‑parameters of the network shown in Fig. 16.16.
Fig. 16.16
Solution From Eqs (16.9) and (16.10), we have
h11 =
V1
I
; h21 = 2
I1 V =0
I1
2
; h12 =
V2 =0
V1
V2
; h22 =
I1 =0
I2
V2
I1 =0
If the port b-b9 is short-circuited, V2 5 0. The circuit is shown in Fig. 16.17 (a).
V
h11 = 1
; V1 = I1Zeq
I1 V =0
2
Fig. 16.17 (a)
The equivalent impedance as viewed from the port a-a9 is 2 V.
∴ V1 5 I1 2 V
h11 =
V1
=2V
I1
726
Circuits and Networks
h21 =
I2
I1
when V2 = 0; − I 2 =
V2 =0
I1
2
1
2
If the port a-a9 is let open, I1 5 0. The circuit is shown in Fig. 16.17 (b).
∴
h21 = −
Fig. 16.17 (b)
Then,
h12 =
V1
V2
I1 =0
V1 = I y 2; I y =
V2 = I X 4; I X =
∴
h12 =
h22 =
16.7
V1
V2
I2
V2
I1 =0
I2
2
=
1
2
=
1

2
I1 =0
I2
2
INVERSE HYBRID (g) PARAMETERS
LO 4
Another set of hybrid matrix parameters can be defined in a similar way as was done in Section 16.6. This
time the current at the input port I1 and the voltage at the output port V2 can be expressed in terms of I2
and V1. The equations are as follows.
I1 5 g11 V1 1 g12 I2
(16.11)
V2 5 g21 V1 1 g22 I2
(16.12)
The coefficients in the above equations are called the inverse hybrid parameters. In matrix notation,
 I1   g11 g12  V1 
 =
 
V2   g21 g22   I 2 
  
 
727
Two-Port Networks
−1
h
g
h 
g12 

It can be verified that  11 12  =  11
 h21 h22 
 g21 g22 




The individual g-parameters may be defined by letting I2 5 0 and V1 5 0 in Eqs (16.11) and (16.12).
Thus, when I2 5 0
g11 =
g21 =
I1
V1
I 2 =0
V2
V1
I 2 =0
 1 
5 Open-circuit input admittance =  
 Z11 
5 Open-circuit voltage gain
When V1 5 0,
g12 =
g22 =
I1
I2
V2
I2
5 Short-circuit reverse current gain
V1 =0
V1 =0
 1 
5 Short-circuit output impedance =  
Y22 
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 4
rrr 16-4.1
rrr 16-4.2
For the two-port network shown in Fig. Q.1, find the h-parameters and the inverse h-parameters.
For the hybrid equivalent circuit shown in Fig. Q.2, determine (a) the input impedance, and (b)
the output impedance.
Fig. Q.1
rrr16-4.3
Fig. Q.2
For the network shown in Fig. Q.3, determine h-parameters at v 5 108 rad/s.
Fig. Q.3
rrr 16-4.4
Using PSpice, find hybrid parameters of the network shown in Fig. Q.4.
728
Circuits and Networks
Fig. Q.4
Frequently Asked Questions linked to LO 4
rrr16-4.1
Determine the Z-ABCD-and h-parameters of the given network as shown in Fig. Q.1 [BPUT 2008]
2I2
2I1
I1
I1
1 ohm
1 ohm
V1
V2
1 ohm
Fig. Q.1
rrr 16-4.2
Explain hybrid parameters for two-port networks and state where one makes use of these
parameters.
[GTU Dec. 2010]
rrr 16-4.3 Obtain hybrid parameters of the interconnected ‘two’ 2-port networks.
[MU 2014]
10 W
8W
VX
4W
3VX
10 W
4W
6W
Fig. Q.3
rrr 16-4.4
Explain the concepts of reciprocity and symmetry. Derive the above conditions for h and ABCD
parameters.
[PTU 2009-10]
rrr 16-4.5 Find the h-parameters of the network shown in Fig. Q.5.
[PTU 2009-10]
rrr 16-4.6 Determine hybrid-parameters for the network shown in Fig. Q.6.
[PU 2010]
I1
2 I1
1W
2W
I1
1W
V1
1F
1W
Fig. Q.5
V1
V2
I1
2W
I2
V2
4W
I3
Fig. Q.6
I2
729
Two-Port Networks
rrr16-4.7
[PU 2012]
Find the current transfer ratio i2/i1 for the network shown in Fig. Q. 7.
2Ia
2W
1
2
I2
I1
1
1W
Ia
1W
V1
I1/2
1
V2
1W
1
2
Fig. Q.7
rrr 16-4.8
16.8
[RGTU June 2014]
Define hybrid parameters.
INTER-RELATIONSHIPS OF DIFFERENT PARAMETERS
16.8.1 Expression of Z-Parameters in Terms of Y-Parameters and Vice
Versa
LO 5
From Eqs (16.1), (16.2), (16.3) and (16.4), it is easy to derive the relation
between the open circuit impedance parameters and the short-circuit admittance
parameters by means of two matrix equations of the respective parameters. By
solving Eqs (16.1) and (16.2) for I1 and I2, we get
I1 =
V1
V2
Z12
Z 22
D z ; and I 2 =
Z11 V1
V21 V2
Dz
where Dz is the determinant of the Z-matrix
Z
D z =  11
 Z 21

Z12 

Z 22 
I1 =
Z 22
Z
V1 − 12 V2
Dz
Dz
(16.13)
I2 =
−Z 21
Z
V1 + 11 V2
Dz
Dz
(16.14)
Comparing Eqs (16.13) and (16.14) with Eqs (16.3) and (16.4) we have
Z
−Z12
Y11 = 22 ; Y12 =
Dz
Dz
Y21 =
Z 21
Z
; Y22 = 11
Dz
Dz
In a similar manner, the Z-parameters may be expressed in terms of the admittance parameters by solving
Eqs (16.3) and (16.4) for V1 and V2.
V1 =
I1 Y12
I 2 Y22
D y ; and V2 =
Y11
Y21
I1
I2
Dy
730
Circuits and Networks
where Dy is the determinant of the Y-matrix
Dy =
Y11
Y12
Y21 Y22
V1 =
Y22
Y
I1 − 12 I 2
Dy
Dy
(16.15)
V2 =
−Y21
Y
I1 + 11 I 2
Dy
Dy
(16.16)
Comparing Eqs (16.15) and (16.16) with Eqs (16.1) and (16.2), we obtain
Y
−Y
Z11 = 22 ; Z12 = 12
Dy
Dy
Z 21 =
−Y21
Y
; Z 22 = 11
Dy
Dy
EXAMPLE 16.5
For a given, Z11 5 3 V; Z12 5 1 V; Z21 5 2 V, and Z22 5 1 V, find the admittance matrix, and the product
of Dy and Dz.
 Z 22 −Z12 


Y11 Y12   ∆ z
∆z 

=
Solution The admittance matrix 5 
Y21 Y22   −Z
Z11 

  21

∆ z 
 ∆ z
 3 1

given Z = 
 2 1


∴ Dz 5 3 – 2 5 1
−1 −1
 =1
∴ Dy = 
−2 3 


(Dy) (Dz) 5 1
16.8.2 General Circuit Parameters or ABCD-Parameters in Terms of Z-Parameters and Y-Parameters
We know that
V1 5 AV2 – BI2; V1 5 Z11 I1 1 Z12 I2; I1 5 Y11 V1 1 Y12 V2
I1 5 CV2 – DI2; V2 5 Z21 I1 1 Z22 I2; I2 5 Y21 V1 1 Y22 V2
A=
V1
V2
;C =
I 2 =0
I1
V2
;B=
I 2 =0
−V1
−I
;D= 1
I 2 V =0
I 2 V =0
2
2
Substituting the condition I2 5 0 in Eqs (16.1) and (16.2), we get
V1
V2
=
I 2 =0
Z11
=A
Z 21
731
Two-Port Networks
Substituting the condition I2 5 0 in Eq. (16.4), we get
V1
V2
=
I 2 =0
−Y22
=A
Y21
Substituting the condition I2 5 0 in (Eq. 16.2), we get
I1
1
=
=C
V2 I =0 Z 21
2
Substituting the condition I2 5 0 in Eqs (16.3) and (16.4), and solving for V2 gives
where Dy is the determinant of the admittance matrix.
I1
V2
=
I 2 =0
−D y
Y21
−I1Y21
Dy
=C
Substituting the condition V2 5 0 in Eq. (16.4), we get
V1
I2
=−
V2 =0
1
=B
Y21
Substituting the condition V2 5 0 in Eqs (16.1) and (16.2) and solving for I 2 =
V
D
− 1
= z =B
I 2 V =0 Z 21
2
where Dz is the determinant of the impedance matrix.
Substituting V2 5 0 in Eq. (16.2), we get
−I1
Z
= 22 = D
I 2 V =0 Z 21
2
Substituting V2 5 0 in Eqs (16.3) and (16.4), we get
−I1
−Y
= 11 = D
I 2 V =0
Y21
2
The determinant of the transmission matrix is given by
– AD 1 BC
Substituting the impedance parameters in A, B, C, and D, we have
BC − AD =
=
BC − AD =
Dz 1
Z Z
− 11 22
Z 21 Z 21 Z 21 Z 21
Dz
2
( Z 21 )
−
Z11Z 22
2
( Z 21 )
−Z12
Z 21
For a bilateral network, Z12 5 Z21
∴ BC – AD 5 – 1
or AD – BC 5 1
−V1Z 21
, we get
Dz
732
Circuits and Networks
Therefore, in a two-port bilateral network, if three transmission parameters are known, the fourth may be
found from the equation AD – BC 5 1.
In a similar manner, the h-parameters may be expressed in terms of the admittance parameters,
impedance parameters, or transmission parameters. Transformations of this nature are possible between
any of the various parameters. Each parameter has its own utility. However, we often find that it is
necessary to convert from one set of parameters to another. Transformations between different parameters,
and the condition under which the two-port network is reciprocal are given in Table 16.1.
Table 16.1 Reciprocality condition for a two-port network
Z
Y
ABCD
A' B' C' D'
h
g
Z11 Z12
Y22 −Y12
∆y ∆y
A ∆T
C C
D9 1
C9 C9
∆h h22
h22 h22
1 −g12
g11 g11
Z21 Z22
−Y21 Y11
∆y ∆y
1 D
CC
∆T 9 A9
C9 C9
−h21 1
h22 h22
g21 ∆ g
g11 g11
Z 22 −Z12
∆z ∆z
Y11 Y12
A9 −1
B9 B9
−Z 21 −Z11
∆z ∆z
Y21 Y22
D −∆T
B B
−1 A
−1 A
B B
1 −h12
h11 h11
∆
hh ∆
∆ g g12
g22 g22
g21 1
−g
g22 g22
Z11 ∆ z
Z21 Z21
1 Z
1 Z22
Z21 Z21
−Y22 −1
Y21 Y21
Y Y
∆Y −Y11
Y21 Y21
Z
Y
AB
CD
AB
CD
−∆T 9 D9
B9 B9
21
h
h11 h11
D9 B9
∆T 9 ∆T 9
C A
C9 A9
∆T 9 ∆T 9
∆h h11
h21 h21
h
−h22 −1
h21 h21
A9 B9
1 h11
h12 h12
g21 g21
−∆ g −g22
g12 g12
1 g22
g21 g21
∆
g11 ∆ g
A9 B9
Z 22 ∆ z
Z12 Z12
−Y11 −1
Y12 Y12
D B
∆T ∆T
C9 D9
1 Z11
Z12 Z12
−∆Y −Y22
Y12 Y12
C A
∆T ∆T
C9 D9
h22 ∆h
h12 h12
−g11 −1
g12 g12
∆ z Z12
Z 22 Z 22
1 −Y12
Y11 Y11
g22 −g12
∆g ∆g
Y21 ∆Y
Y11 Y11
B9 1
A9 A9
∆T 9 C9
A9 A9
h11 h12
−Z 21 1
Z 22 Z 22
B ∆T
D D
−1 C
D D
h21 h22
−g21 g11
∆g ∆g
1 −Z12
Z11 Z11
∆Y Y12
Y22 Y22
g11 g12
−Y21 1
Y22 Y22
C 9 −1
D9 D9
∆T 9 B9
D9 D9
h22 −h12
∆h ∆h
Z 21 ∆ z
Z11 Z11
C −∆T
A A
1 B
AA
Z12 5 Z21
Y12 5 Y21
h
g
The two-port is
reciprocal if
The determinant of the
transmission
matrix 5 1
(DT 5 1)
The determinant of
the inverse
transmission
matrix 5 1
−h21 h11
∆h ∆h
g21 g22
h12 5 – h21
g12 5 – g21
Two-Port Networks
733
EXAMPLE 16.6
The impedance parameters of a two‑port network are Z11 5 6 V; Z22 5 4 V; Z12 5 Z21 5 3 V. Compute the
Y‑parameters and ABCD‑ parameters and write the describing equations.
Solution ABCD-parameters are given by
A=
Z11 6
Z Z − Z12 Z 21
= = 2; B = 11 22
=5 V
Z 21 3
Z 21
C=
Z
1
1
4
= ; D = 22 =
Z 21 3
Z 21 3
Y-parameters are given by
Z 22
−Z12
4
−1
Y11 =
; Y12 =

=
=
Z11Z 22 − Z12 Z 21 15
Z11Z 22 − Z12 Z 21
5
Y21 = Y12 =
−Z12 −1
;
=
∆Z
5
Y22 =
Z11
2
= 
Z11Z 22 − Z12 Z 21 5
The equations, using Z, parameters, are
V1 5 6I1 1 3I2
V2 5 3I1 1 4I2
Using Y‑parameters,
4
1
I1 = V1 − V2
15
5
2
−1
V1 + V2
5
5
Using ABCD-parameters,
V1 5 2V2 – 5I2
1
4
I1 = V2 − I 2
3
3
I2 =
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 5
rrr 16-5.1 The hybrid parameters of a two-port network shown in Fig. Q.1 are h11 5 1.5 K; h12 5 2 3 10–3; h21 5 250;
h22 5 150 3 10–6
(a) Find V2 (b). Draw the Z‑ parameter equivalent circuit.
Fig. Q.1
734
Circuits and Networks
Frequently Asked Questions linked to LO 5
ABCD
r
y
r
r
Z
rr
Z
Z
Z
Z
ABCD
16.9
INTERCONNECTION OF TWO-PORT NETWORKS
16.9.1 Series Connection of a Two-port Network
It has already been shown in Section 16.4.1 that when two-port networks are
connected in cascade, the parameters of the interconnected network can be
conveniently expressed with the help of ABCD-parameters. In a similar way, the
Z-parameters can be used to describe the
parameters of series-connected two-port
networks; and Y-parameters can be used to
describe parameters of parallel connected
two-port networks. A series connection of
two-port networks is shown in Fig. 16.18.
Let us consider two two-port networks,
connected in series as shown. If each port
has a common reference node for its input
and output, and if these references are
connected together then the equations
of the networks X and Y in terms of Z‑
parameters are
Fig. 16.18
V1X 5 Z11X I1X 1 Z12X I2X
V2X 5 Z21X I1X 1 Z22X I2X
V1Y 5 Z11Y I1Y 1 Z12Y I2Y
V2Y 5 Z21Y I1Y 1 Z22Y I2Y
From the interconnection of the networks, it is clear that
I1 5 I1X 5 I1Y ; I2 5 I2X 5 I2Y
and V1 5 V1X 1 V1Y ; V2 5 V2X 1 V2Y
∴ V1 5 Z11X I1 1 Z12X I2 1 Z11Y I1 1 Z12Y I2
5 (Z11X 1 Z11Y)I1 1 (Z12X 1 Z12Y) I2
V2 5 Z21X I1 1 Z22X I2 1 Z21Y I1 1 Z22Y I2
5 (Z21X 1 X21Y)I1 1 (Z22X 1 Z22Y)I2
LO 6
Two-Port Networks
735
The describing equations for the series-connected two-port network are
V1 5 Z11 I1 1 Z12 I2
V2 5 Z21 I1 1 Z22 I2
where Z11 5 Z11X 1 Z11Y; Z12 5 Z12X 1 Z12Y
Z21 5 Z21X 1 Z21Y; Z22 5 Z22X 1 Z22Y
Thus, we see that each Z-parameter of the series network is given as the sum of the corresponding
parameters of the individual networks.
16.9.2 Parallel Connection of Two Two-port Networks
Let us consider two two-port networks connected in parallel as shown in Fig. 16.19. If each two-port has a
reference node that is common to its input and output port, and if the two ports are connected so that they have
a common reference node, then the equations of the networks X and Y in terms of Y-parameters are given by
Fig. 16.19
I1X 5 Y11X V1X 1 Y12X V2X
I2X 5 Y21X V1X 1 Y22X V2X
I1Y 5 Y11Y V1Y 1 Y12Y V2Y
I2Y 5 Y21Y V1Y 1 Y22Y V2Y
From the interconnection of the networks, it is clear that
V1 5 V1X 5 V1Y; V2 5 V2X 5 V2Y
and I1 5 I1X 1 I1Y; I2 5 I2X 1 I2Y
∴ I1 5 Y11X V1 1 Y12X V2 1 Y11Y V1 1 Y12Y V2
5 (Y11X 1 Y11Y) V1 1 (Y12X 1 Y12Y) V2
I2 5 Y21X V1 1 Y22X V2 1 Y21Y V1 1 Y22Y V2
5 (Y21X 1 Y21Y) V1 1 (Y22X 1 Y22Y) V2
The describing equations for the parallel connected two-port networks are
I1 5 Y11 V1 1 Y12 V2
I2 5 Y21 V1 1 Y22 V2
736
Circuits and Networks
where Y11 5 Y11X 1 Y11Y; Y12 5 Y12X 1 Y12Y
Y21 5 Y21X 1 Y21Y; Y22 5 Y22X 1 Y22Y
Thus, we see that each Y-parameter of the parallel network is given as the sum of the corresponding
parameters of the individual networks.
EXAMPLE 16.7
Two networks shown in Figs 16.20 (a) and (b) are connected in series. Obtain the Z‑parameters of the
combination. Also verify by direct calculation.
Fig. 16.20
Solution The Z-parameters of the network in Fig. 16.20 (a) are
Z11X 5 3 V Z12X 5 Z21X 5 2 V Z22X 5 3 V
The Z-parameters of the network in Fig. 16.20 (b) are
Z11Y 5 15 V Z21Y 5 5 V Z22Y 5 25 V Z12Y 5 5 V
The Z-parameters of the combined network are
Z11 5 Z11X 1 Z11Y 5 18 V
Z12 5 Z12X 1 Z12Y 5 7 V
Z21 5 Z21X 1 Z21Y 5 7 V
Z22 5 Z22X 1 Z22Y 5 28 V
Check
If the two networks are connected in series as shown in Fig. 16.20 (c), the Z-parameters are
Z11 =
Z 21 =
Z 22 =
Fig. 16.20 (c)
Z12 =
V1
I1
I 2 =0
V2
I1
I 2 =0
V2
I2
I1 =0
V1
I2
I1 =0
= 18 Ω
=7Ω
= 28 Ω
=7Ω
737
Two-Port Networks
EXAMPLE 16.8
Two identical sections of the network shown in Fig. 16.21 are connected in parallel. Obtain the Y‑parameters
of the combination.
Fig. 16.21
Solution The Y-parameters of the network in Fig. 16.21 are (see Example 16.2)
Y11 =
1
5
−1
−1
 Y21 =
 Y22 =  Y12 =

2
4
8
4
If two such networks are connected in parallel then the Y-parameters of the combined network are
Y11 =
1 1
−1
−1
+ = 1  Y21 =
×2 =

2 2
4
2
5
5
−1
−1
Y22 = × 2 =  Y12 =

×2 =
8
4
4
2
16.10
T- AND p-REPRESENTATIONS
A two-port network with any number of elements may be converted into a twoLO 7
port three-element network. Thus, a two-port network may be represented by an
p
equivalent T-network, i.e. three impedances are connected together in the form of
a T as shown in Fig. 16.22.
It is possible to express the elements of the T-network in terms of Z-parameters, or ABCD-parameters as
explained below.
Z-parameters of the network
Z11 =
Z 21 =
Fig. 16.22
Z 22 =
Z12 =
V1
I1
I 2 =0
V2
I1
I 2 =0
V2
I2
I1 =0
V1
I2
I1 =0
= Z a + Zc
= Zc
= Zb + Zc
= Zc
738
Circuits and Networks
From the above relations, it is clear that
Za 5 Z11 – Z21
Zb 5 Z22 – Z12
Zc 5 Z12 5 Z21
ABCD‑parameters of the network
A=
B=
V1
V2
=
I 2 =0
Z a + Zc
Zc
−V1
I 2 V =0
2
When 2-29 is short-circuited,
−I 2 =
V1Zc
Zb Zc + Z a ( Zb + Zc )
B = ( Z a + Zb ) +
C=
D=
I1
V2
=
I 2 =0
Z a Zb
Zc
1
Zc
−I1
I 2 V =0
2
When 2-29 is short-circuited,
−I 2 = I1
D=
Zc
Zb + Zc
Zb + Zc
Zc
From the above relations, we can obtain
Za =
1
A −1
D −1
; Zb =
; Zc =
C
C
C
EXAMPLE 16.9
The Z‑parameters of a two‑port network are Z11 5 10 V; Z22 5 15 V; Z12 5 Z21 5 5 V. Find the equivalent
T‑network and ABCD‑parameters.
Solution The equivalent T-network is shown in Fig. 16.23,
where
and
Za 5 Z11 – Z21 5 5 V
Zb 5 Z22 – Z12 5 10 V
Zc 5 5 V
739
Two-Port Networks
The ABCD-parameters of the network are
Z
Z Z
A = a + 1 = 2; B = ( Z a + Zb ) + a b = 25 Ω
Zc
Zc
C=
Z
1
= 0.2  D = 1 + b = 3
Zc
Zc
In a similar way, a two-port network may be represented by an equivalent p-network, i.e. three
impedances or admittances are connected together in the form of p as shown in Fig. 16.24.
Fig. 16.23
Fig. 16.24
It is possible to express the elements of the p-network in terms of Y-parameters or ABCD‑parameters
as explained below.
Y‑parameters of the network are
Y11 =
I1
= Y1 + Y2
V1 V =0
2
Y21 =
Y22 =
Y12 =
I2
V1
V2 =0
I2
V2
V1 =0
I1
V2
V1 =0
= −Y2
= Y3 + Y2
= −Y2
From the above relations, it is clear that
Y1 5 Y11 1 Y21
Y2 5 – Y12
Y3 5 Y22 1 Y21
Writing ABCD-parameters in terms of Y-parameters yields the following results.
A=
−Y22 Y3 + Y2
=
Y21
Y2
B=
1
−1
=
Y21 Y2
740
Circuits and Networks
C=
YY
−∆y
= Y1 + Y3 + 1 3
Y21
Y2
D=
−Y11 Y1 + Y2
=
Y2
Y21
From the above results, we can obtain
D −1
B
1
Y2 =
B
A −1
Y3 =
B
Y1 =
EXAMPLE 16.10
The port currents of a two‑port network are given by
I1 5 2.5V1 – V2
I2 5 – V1 1 5V2
Find the equivalent p‑network.
Solution Let us first find the Y-parameters of the network.
Y11 =
I1
I
= 2.5 ; Y21 = 2
V1 V =0
V1
V2 =0
I1
V2
V1 =0
2
Y12 =
= −1 ; Y22 =
V1 =0
I2
V2
= −1 
=5
The equivalent p-network is shown in Fig. 16.25.
Fig. 16.25
where Y1 5 Y11 1 Y21 5 1.5  ;
Y2 5 – Y12 5 – 1
and Y3 5 Y22 1 Y12 5 4 
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 7
rrr 16-7.1
Obtain a -equivalent circuit for Fig. Q.1.
I2
Fig. Q.1
741
Two-Port Networks
Frequently Asked Questions linked to LO 7
rrr
T
6W
1W
4W
2W
Input
Output
Fig. Q.1
16.11
TERMINATED TWO-PORT NETWORK
16.11.1 Driving-Point Impedance at the Input Port of a Load-Terminated Network
Figure 16.26 shows a two-port network connected to an ideal generator
at the input port and to a load impedance at the output port. The input
impedance of this network can be expressed in terms of parameters of the
two-port network.
LO 8
Fig. 16.26
In Terms of Z-Parameters
The load at the output port 2-29 imposes the following constraints on the port voltage and current,
i.e., V2 5 – ZL I2
Recalling Eqs (16.1) and (16.2), we have
V1 5 Z11 I1 1 Z12 I2
V2 5 Z21 I1 1 Z22 I2
742
Circuits and Networks
Substituting the value of V2 in Eq. (16.2), we have
– ZL I2 5 Z21 I1 1 Z22 I2
from which
I2 =
−I1Z 21
Z L + Z 22
Substituting the value of I2 in Eq. (16.1) gives
V1 = Z11 I1 −
Z12 Z 21 I1
Z L + Z 22

Z Z 
V1 = I1  Z11 − 12 21 

Z L + Z 22 
Hence, the driving-point impedance at 1-19 is
V1
Z Z
= Z11 − 12 21
I1
Z L + Z 22
If the output port is open, i.e. ZL → `, the input impedance is given by V1/I1 5 Z11
If the output port is short-circuited, i.e. ZL → 0,
The short-circuit driving-point impedance is given by
Z11Z 22 − Z12 Z 21
1
=
Z 22
Y11
In Terms of Y-Parameters
If a load admittance YL is connected across the output port, the constraint imposed on the output port voltage
and current is
1
−I 2 = V2YL , where YL =
ZL
Recalling Eqs (16.3) and (16.4), we have
I1 5 Y11 V1 1 Y12 V2
I2 5 Y21 V1 1 Y22 V2
Substituting the value of I2 in Eq. (16.4), we have
– V2 YL 5 Y21 V1 1 Y22 V2
 Y21 
V2 = −
V
Y + Y  1
L
22
Substituting the value of V2 in Eq. (16.3), we have
I1 = Y11V1 −
From which
Y12Y21V1
YL + Y22
Y Y
I1
= Y11 − 12 21
YL + Y22
V1
743
Two-Port Networks
Hence, the driving-point impedance is given by
V1
Y22 + YL
=
I1 Y11 (Y1 + Y22 ) − Y12Y21
If the output port is open, i.e., YL → 0
V1 Y22
=
= Z11
I1 D y
If the output port is short-circuited, i.e. YL → `
Then Yin 5 Y11
In a similar way, the input impedance of the load-terminated two-port network may be expressed in terms
of other parameters by simple mathematical manipulations. The results are given in Table 16.2.
Table 16.2 Output impedance
In terms of
Driving- point
impedance
at the input
port, or input
impedance
V

 I



Z parameters
Y parameters
ABCD
A9B9C9D9
h parameter
g parameter
∆ z + Z11Z L
Z22 + Z L
Y22 + YL
∆ y + Y11YL
AZ L + B
CZ L + D
B ′ − D ′Z L
C ′Z L − A′
∆h Z L + h11
1 + h22 Z L
1 + g22YL
∆ gYL + g11
∆ z + Z22 Z s
Z1 + Z11
Y11 + Ys
∆ y + YsY22
DZ s + B
CZ s + A
A′ Z s + B ′
C ′Z s + D ′
h11 + Z s
∆h + h22 Z s
g22 + ∆s
1 + g11Z s
Driving- point
impedance
at the output
port, or output
impedance
V

 I



Note The above relations are obtained when Vs = 0 and Is = 0 at the input port.
16.11.2 Driving-Point Impedance at the Output Port with Source Impedance at the Input Port
Let us consider a two-port network connected to a generator at the input port with a source impedance Zs
as shown in Fig. 16.27. The output impedance, or the driving- point impedance, at the output port can be
evaluated in terms of the parameters of the two-port network.
Fig. 16.27
744
Circuits and Networks
In terms of Z-parameters If I1 is the current due to Vs at the port 1-19, from Eqs (16.1) and (16.2),
we have
V2 5 Z21I1 1 Z22I2
V1 5 Vs – I1Zs
5 Z11 I1 1 Z12I2 – (I1) (Zs 1 Z11) 5 Z12I2 – Vs
Z12 I 2 −Vs
Z s + Z11
−I1 =
Substituting I1 in Eq. (16.2), we get
V2 = −Z 21
( Z12 I 2 −Vs )
Z s + Z11
+ Z 22 I 2
With no source voltage at the port 1-19, i.e. if the source Vs is short-circuited,
V2 =
−Z 21Z12
I 2 + Z 22 I 2
Z s + Z11
Hence, the driving-point impedance at the port 2-29 5 V2
I2
V2 Z 22 Z s + Z 22 Z11 − Z 21Z12
=
I2
Z s + Z11
or
∆ z + Z 22 Z s
Z s + Z11
If the input port is open, i.e. Zs → `
Then
 ∆z


+ Z 22 

V2  Z s

=


Z
I2
11
 1+


Z s 
= Z 22
Z s =∞
If the source impedance is zero with a short-circuited input port, the driving-point impedance at the output
port is given by
V2 ∆ z
1
=
=
I2
Z11 Y22
In terms of Y-parameters Let us consider a two-port network connected to a current source at the
input port with a source admittance Ys as shown in Fig. 16.28.
At the port 1-19,
I1 5 Is – V1 Ys
Recalling Eqs (16.3) and (16.4), we have
I1 5 Y11 V1 1 Y12 V2
I2 5 Y21 V1 1 Y22 V2
Substituting I1 in Eq. (16.3), we get
Is – V1Ys 5 Y11V1 1 Y12V2
745
Two-Port Networks
Fig. 16.28
–V1(Ys 1 Y11) 5 Y12 V2 – Is
−V1 =
Y12V2 − I s
Ys + Y11
Substituting V1 in Eq. (16.4), we get
Y V − I s 
I 2 = −Y21  12 2
+Y V
 Y + Y  22 2
s
11
With no source current at 1-19, i.e. if the current source is open-circuited,
I2 =
−Y21Y12V2
+ Y22V2
Ys + Y11
Hence, the driving-point admittance at the output port is given by
I 2 Y22Ys + Y22Y11 − Y21Y12
=
V2
Ys + Y11
or
D y + Y22Ys
Ys + Y11
If the source admittance is zero, with an open-circuited input port, the driving-point admittance at the output
port is given by
I2 D y
1
=
=
= Y22
V2 Y11 Z 22
In a similar way, the output impedance may be expressed in terms of the other two-port parameters by simple
mathematical manipulations. The results are given in Table 16.2.
EXAMPLE 16.11
Calculate the input impedance of the network shown in Fig. 16.29.
Fig. 16.29
746
Circuits and Networks
Solution Let us calculate the input impedance in terms of Z-parameters. The Z-parameters of the given
network (see Solved Problem 16.1) are Z11 5 2.5 V; Z21 5 1 V; Z22 5 2 V; Z12 5 1 V
From Section 16.11.1, we have the relation
V1
Z Z
= Z11 − 12 21
I1
Z L + Z 22
where ZL is the load impedance 5 2 V
V1
1
= 2.25 Ω
= 2.5 −
I1
2+2
The source resistance is 1 V.
∴ Zin 5 1 1 2.25 5 3.25 V
EXAMPLE 16.12
Calculate the output impedance of the network shown in Fig. 16.30 with a source admittance of 1  at the
input port.
Fig. 16.30
Solution Let us calculate the output impedance in terms of Y-parameters. The Y-parameters of the given
network (see Example 16.2) are
Y11 =
1
5
−1
; Y22 = ; Y21 = Y12 =

2
8
4
From Section 16.11.2, we have the relation
I 2 Y22Ys + Y22Y11 − Y21Y12
=
V2
Ys + Y11
where Ys is the source admittance 5 1 mho
5 1 1
5
×1 + × −
I2
8 2 16 = 7 
=8
Y22 =
1
12
V2
1+
2
or
Z 22 =
12

7
Two-Port Networks
16.12
747
LATTICE NETWORKS
One of the common four-terminal two-port networks is the lattice, or bridge
LO 9
network shown in Fig. 16.31 (a). Lattice networks are used in filter sections and
are also used as attenuaters. Lattice structures are sometimes used in preference
to ladder structures in some special applications. Za and Zd are called series arms, and Zb and Zc are called the
diagonal arms. It can be observed that, if Zd is zero, the lattice structure becomes a p-section. The lattice network
is redrawn as a bridge network as shown in Fig. 16.31 (b).
Fig. 16.31
16.12.1 Z-Parameters
Z11 =
V1
I1
When I2 5 0; V1 = I1
∴ Z11 =
I 2 =0
( Z a + Zb )( Z d + Zc )
Z a + Zb + Zc + Z d
( Z a + Z b )( Z d + Z c )
Z a + Zb + Zc + Z d
If the network is symmetric, then Za 5 Zd and Zb 5 Zc
Z + Zb
∴ Z11 = a
2
V
Z 21 = 2
I1 I =0
2
When I2 5 0, V2 is the voltage across 2229.
 Zb
Zd 

V2 = V1 
−
 Z a + Z b Z c + Z d 
Substituting the value of V1 from Eq. (16.17), we have
 I ( Z + Z b )( Z d + Z c )   Z b ( Z c + Z d ) − Z d ( Z a + Z b ) 


V2 =  1 a

( Z a + Zb )( Zc + Z d ) 
 Z a + Zb + Zc + Z d  
Zb Zc − Z a Z d
V2 Zb ( Zc + Z d ) − Z d ( Z a + Zb )
=
=
I1
Z a + Zb + Zc + Z d
Z a + Zb + Zc + Z d
∴ Z 21 =
Zb Zc − Z a Z d
Z a + Zb + Zc + Z d
(16.17)
748
Circuits and Networks
If the network is symmetric, Za 5 Zd, Zb 5 Zc
Z 21 =
Zb − Z a
2
When the input port is open, I1 5 0
V
Z12 = 1
I 2 I =0
1
The network can be redrawn as shown in Fig. 16.31 (c).
 Zc
Zd 

V1 = V2 
−
 Z a + Z c Z b + Z d 
(16.18)
 ( Z + Z c )( Z d + Z b )

V = I  a

 Z a + Zb + Zc + Z d 
(16.19)
Substituting the value of V2 in Eq. (16.18), we get
 Z ( Z + Z d ) − Z d ( Z a + Z c )

V1 = I 2  c b

Z a + Zb + Zc + Z d


Zc Zb − Z a Z d
V1
=
I 2 Z a + Zb + Zc + Z d
If the network is symmetric, Za 5 Zd; Zb 5 Zc
Z b2 − Z a2
V1
=
I 2 2 (Z a + Zb )
∴
Fig. 16.31 (c)
Z12 =
Zb − Z a
2
Z 22 =
V2
I2
I 2 =0
From Eq. (16.19), we have
V2 ( Z a + Zc ) − ( Z d + Zb )
=
I2
Z a + Zb + Zc + Z d
If the network is symmetric,
Za 5 Zd; Zb 5 Zc
Z 22 =
Z a + Zb
= Z11
2
749
Two-Port Networks
From the above equations, Z11 = Z 22 =
and
∴
Z a + Zb
2
Zb − Z a
2
Zb 5 Z11 1 Z12
Z12 = Z 21 =
Za 5 Z11 – Z12.
EXAMPLE 16.13
Obtain the lattice equivalent of a symmetrical T‑network shown in Fig. 16.32.
Fig. 16.32
A two-port network can be realised as a symmetric lattice if it is reciprocal and symmetric. The
Z-parameters of the network are (see Example 16.1). Z11 5 3 V; Z12 5 Z21 5 2 V; Z22 5 3 V.
Since Z11 5 Z22; Z12 5 Z21, the given network is symmetrical and
reciprocal
∴ the parameters of the lattice network are
Solution
Za 5 Z11 – Z12 5 1 V
Zb 5 Z11 1 Z12 5 5 V
The lattice network is shown in Fig. 16.33.
Fig. 16.33
EXAMPLE 16.14
Obtain the lattice equivalent of a symmetric p‑network shown in Fig. 16.34.
Fig. 16.34
750
Circuits and Networks
Solution The Z-parameters of the given network are
Z11 5 6 V 5 Z22; Z12 5 Z21 5 4 V
Hence, the parameters of the lattice network are
Za 5 Z11 – Z12 5 2 V
Zb 5 Z11 1 Z12 5 10 V
The lattice network is shown in Fig.16.35
16.13
Fig. 16.35
IMAGE PARAMETERS
The image impedance ZI1 and ZI2 of a two-port network shown in Fig. 16.36 are
two values of impedance such that, if the port 1-19 of the network is terminated
in ZI1, the input impedance of the port 2-29 is ZI2; and if the port 2-29 is
terminated in ZI2, the input impedance at the port 1-19 is ZI1.
LO 10
Fig. 16.36
Then, ZI1 and ZI2 are called image impedances of the two-port network shown in Fig. 16.36. These
parameters can be obtained in terms of two-port parameters. Recalling Eqs (16.5) and (16.6) in Section 16.4,
we have
V1 5 AV2 – BI2
I1 5 CV2 – DI2
Fig. 16.37
If the network is terminated in ZI2 at 2-29 as shown in Fig. 16.37,
V2 5 –I2 ZI2
Two-Port Networks
V1
AV2 − BI 2
=
= Z I1
I1 CV2 − DI 2
or
Z I1 =
− AI 2 Z I 2 − BI 2
−CI 2 Z I 2 − DI 2
Z I1 =
− AZ I 2 − B
−CZ I 2 − D
Z I1 =
AZ I 2 + B
CZ I 2 + D
Fig. 16.38
Similarly, if the network is terminated in ZI1 at the port 1-19 as shown in Fig. 16.38, then
V1 5 – I1ZI1
V
= ZI
I
∴
from which,
−Z I1 =
V1
AV2 − BI 2
=
I1 CV2 − DI 2
−Z I 1 =
AI 2 Z I 2 − BI 2
CI 2 Z I 2 − DI 2
−Z I 1 =
AZ I 2 − B
CZ I 2 − D
ZI 2 =
DZ I 1 + B
CZ I 1 + A
Substituting the value of ZI1 in the above equation,

 (− AZ I 2 + B )
 − AZ I 2 + B 
+B
Z I 2 C
+ A = D 
 CZ I 2 − D 
 (CZ I 2 − D )



751
752
Circuits and Networks
from which,
ZI =
BD
AC
AB
CD
If the network is symmetrical, then A 5 D
Similarly, we can find Z I =
∴
Z I1 = Z I 2 =
B
C
If the network is symmetrical, the image impedances ZI1 and ZI2 are equal to each other; the image
impedance is then called the characteristic impedance, or the iterative impedance, i.e. if a symmetrical
network is terminated in ZL, its input impedance will also be ZL, or its impedance transformation ratio is
unity. Since a reciprocal symmetric network can be described by two independent parameters, the image
parameters ZI1 and ZI2 are sufficient to characterise reciprocal symmetric networks. ZI1 and ZI2, the two image
parameters, do not completely define a network. A third parameter called image transfer constant f is also
used to describe reciprocal networks. This parameter may be obtained from the voltage and current ratios.
If the image impedance ZI2 is connected across the port 2-29, then
∴
V1 5 AV2 – BI2
(16.20)
V2 5 – I2 ZI2
(16.21)

B 
 V2
V1 =  A +

Z I 2 
(16.22)
I1 5 CV2 – DI2
(16.23)
I1 5 – [CZI2 1 D]I2
(16.24)
From Eq. (16.22),
V1 
B 
AC
 = A+ B
= A+
V2 
Z I 2 
BD
V1
ABCD
= A+
V2
D
(16.25)
From Eq. (16.24),
−I1
BD
= [CZ I 2 + D] = D + C
I2
AC
−I1
ABCD
= D+
I2
A
Multiplying Eqs (16.25) and (16.26), we have
−V1 I1  AD + ABCD  AD + ABCD 


× =


V2 I 2 
D
A
(16.26)
753
Two-Port Networks
2
−V1 I1
× = ( AD + BC )
V2 I 2
or
AD + BC =
−V1 I1
×
V2 I 2
AD + AD −1 =
−V1 I1
×
V2 I 2
Let
(∵ AD − BC = 1)
cos h f = AD ; sin h f = AD −1
tan h f =
∴
Also,
AD −1
AD
f = tan h−1
=
BC
AD
BC
AD
ef = cos h f + sin h f = −
V1 I1
V2 I 2
 VI  1
V I 
f = log e − 1 1  = log e  1 1 
 V2 I 2  2
 V2 I 2 
Since
V1 5 ZI1 I1; V2 5 – I2 ZI2
I 
Z 
1
f = log e  I 1  + log  1 
 Z I 2 
 I 2 
2
For symmetrical reciprocal networks, ZI1 5 ZI2
I 
f = log e  1  = g
 I 2 
where g is called the propagation constant.
EXAMPLE 16.15
Determine the image parameters of the T‑network shown in Fig.16.39.
Solution The ABCD-parameters of the network are
6
17
1
7
A = ; B = ; C = ; D = (See Ex. 16.3)
5
5
5
5
Fig. 16.39
754
Circuits and Networks
Since the network is not symmetrical, f, ZI1, and ZI2 are to be evaluated to describe the network.
AB
=
Z I1 =
CD
6 17
×
5 5 = 3.817 V
1 7
×
5 5
BC
=
AC
17 7
×
5 5 = 4.453 V
6 1
×
5 5
ZI 2 =
f = tan h−1
or
17
BC
= tan h−1
AD
42
f = ln [ AD + AD −1 ]
f 5 0.75
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 10
rrr16-10.1
Obtain the image parameters of the symmetric
lattice network given in Fig. Q.1.
rrr 16-10.2
Determine the Z-parameters and image
parameters of a symmetric lattice network whose
series arm impedance is 10 V and diagonal arm
impedance is 20 V.
Fig. Q.1
Frequently Asked Questions linked to LO 10
rrr
Z
1W
5W
rrr
rrr
2W
Z
rrr
T
Fig. Q.4
Additional Solved Problems
PROBLEM 16.1
Find the Z‑parameters for the circuit shown in Fig. 16.40.
Two-Port Networks
755
Fig. 16.40
Solution
Z11 =
V1
I1
I 2 =0
When I2 5 0; V1 can be expressed in terms of I1 and the equivalent impedance of the circuit looking from the
terminal a-a9 is as shown in Fig. 16.41 (a).
6× 2
Zeq = 1 +
= 2.5 V
6+2
V1 5 I1 Zeq 5 I1 2.5
Z11 =
Z 21 =
Fig. 16.41 (a)
V1
I1
V2
I1
= 2.5 V
I 2 =0
I 2 =0
V2 is the voltage across the 4 V impedance as shown in Fig. 16.41 (b).
Fig. 16.41 (b)
Let the current in the 4 V impedance be Ix.
2 I
I x = I1 × = 1
8 4
V2 = I x 4 =
Z 21 =
Z 22 =
I1
× 4 = I1
4
V2
I2
I 2 =0
V2
I2
I1 =0
= 1Ω
756
Circuits and Networks
When the port a-a9 is open-circuited, the voltage at the port b-b9 can be expressed in terms of I2, and the equivalent
impedance of the circuit viewed from b-b9 is as shown in Fig. 16.41 (c).
Fig. 16.41 (c)
V2 5 I2 3 2
∴ 22 5
Z12 =
V2
I2
52V
I150
V1
I2
I1 =0
V1 is the voltage across the 2 V (parallel) impedance and let the current in the 2 V (parallel) impedance is
IY as shown in Fig. 16.41 (d).
Fig. 16.41 (d)
IY =
I
V1 5 2IY
V1 = 2
∴
Z12 =
I2
2
V1
I2
= 1V
I1 =0
Here, Z12 5 Z21, which indicates the bilateral property of the network. The describing equations for
this two-port network in terms of impedance parameters are
V1 5 2.5I1 1 I2
V2 5 I1 1 2I2
Two-Port Networks
757
PROBLEM 16.2
Find the short‑circuit admittance parameters for the circuit shown in Fig. 16.42.
Fig. 16.42
Solution The elements in the branches of the given two-port network are admittances. The admittance
parameters can be determined by short-circuiting the two-ports.
When the port b-b9 is short-circuited, V2 5 0. This circuit is shown in Fig. 16.43 (a).
V1 5 I1 Zeq
where Zeq is the equivalent impedance as viewed from a-a9.
1
Z eq =
Yeq
Yeq 5 YA 1 YB
V =
I
YA + YB
Fig. 16.43 (a)
Y11 =
I1
= (YA + YB )
V1 V =0
2
With the port b-b9 short-circuited, the nodal equation at the node 1 gives
– I2 5 V1 YB
∴
Y21 =
I2
V1
= −YB
V2 =0
When the port a-a9 is short-circuited; V1 5 0. This circuit is shown in Fig. 16.43 (b).
V2 5 I2 Zeq
where Zeq is the equivalent impedance as viewed from b-b9.
1
Z eq =
Yeq
Yeq 5 Yb 1 Yc
∴
Fig. 16.43 (b)
V =
I
YB + YC
758
Circuits and Networks
Y22 =
I2
V2
= (YB + YC )
V1 =0
With the port a-a9 short-circuited, the nodal equation at the node 2 gives
– I1 5 V2 YB
Y12 =
I1
V2
= −YB
V1 =0
The describing equations in terms of the admittance parameters are
I1 5 (YA 1 YB)V1 – YBV2
I2 5 – YBV1 1 (YC 1 YB)V2
PROBLEM 16.3
Find the Z‑parameters of the RC ladder network shown in Fig. 16.44.
Fig. 16.44
Solution With the port b-b9 open-circuited and assuming mesh currents with V1(S ) as the voltage at a-a9, the
corresponding network is shown in Fig. 16.45 (a).
Fig. 16.45 (a)
The KVL equations are as follows:
V2(S ) 5 I3(S )

1
I 3 ( S )×2 +  = I1 ( S )

S


1
1 +  I1 ( S ) − I 3 ( S ) = V1 ( S )
 S 
(16.27)
(16.28)
(16.29)
Two-Port Networks
From Eq. (16.28),
 S 
I 3 ( S ) = I1 ( S )

1 + 2 S 
From Eq. (16.29),
 S + 1
S

( )
( )
( )
 S  I1 S − I1 S 1 + 2 S = V1 S
759
1 + S
S 
−
I1 ( S )
 = V1 ( S )
 S
1 + 2 S 
 S 2 + 3S + 1
 = V1 ( S )
I1 ( S )
 S (1 + 2 S ) 
Z11 =
Also,
V1 ( S )
I1 ( S ) I
=
( S 2 + 3S + 1)
S (1 + 2 S )
2 =0
V2 ( S ) = I 3 ( S ) = I1 (S )
Z 21 =
V2 ( S )
I1 ( S ) I
=
2 =0
S
1 + 2S
S
1 + 2S
With the port a-a9 open-circuited and assuming mesh currents with V2(S ) as the voltage as b-b9, the
corresponding network is shown in Fig. 16.45 (b).
Fig. 16.45 (b)
The KVL equations are as follows:
V1(S ) 5 I3(S )
(16.30)


2 + 1  I 3 ( S ) = I 2 ( S )

S 
(16.31)
V2(S ) 5 I2(S ) – I3(S )
From Eq. (16.31),
 S 
I 3 ( S ) = I 2 ( S )

 2 S + 1
From Eq. (16.32),
 S 
V2 (S ) = I 2 (S ) − I 2 (S )
 2S + 1
(16.32)
760
Circuits and Networks

S 
V2 ( S ) = I 2 ( S )1−

 2 S + 1
Z 22 =
V2 ( S )
S +1
=
(
)
I 2 S I (S )=0 2 S + 1
1
Also,
 S 
V1 ( S ) = I 3 ( S ) = I 2 ( S )

 2 S + 1
Z12 =
 S 
V1 ( S )
= 

I 2 ( S ) I (S )=0  2 S + 1
1
The describing equations are
 S 2 + 3S + 1 
 I +  S  I
V1 ( S ) = 
 1  2 S + 1 2
 3(2 S + 1) 
 S 
 S +1 
I +
I
V2 (S ) = 
 2S + 1 1  2S + 1 2
PROBLEM 16.4
Find the transmission parameters for the circuit shown in Fig. 16.46.
a
b
a′
b′
Fig. 16.46
Solution Recalling Eqs (16.5) and (16.6), we have
V1 5 AV2 – BI2
I1 5 CV2 – DI2
When the port b-b9 is short- circuited with V1 across
−V
a-a9, V2 5 0, B = 1 and the circuit is as shown in
I2
Fig. 16.47 (a).
V1
I1 = V1
2
∴ B52V
−I
D= 1 =2
I2
−I 2 =
Fig. 16.47 (a)
761
Two-Port Networks
When the port b-b9 is open-circuited with V1 across
a-a9, I2 5 0, A 5 V1/V2 and the circuit is as shown in Fig.
16.47 (b), where V1 is the voltage across the 2 V resistor
across the port a-a9 and V2 is the voltage across the 2 V
resistor across the port b-b9 when I2 5 0.
V
From Fig. 16.47 (b), IY = 1
4
V1
2
V2 = 2× IY =
Fig. 16.47 (b)
A52
Ix =
From Fig. 16.47 (b),
V1
2
C=
I1
V2
where
I1 =
3V1
4
Therefore,
C=
3

2
PROBLEM 16.5
I1
Find the h‑ parameters for the network in Fig. 16.48.
I2
2Ω
V1
4Ω
Solution When V2 5 0, the network is as shown in Fig.
16.49.
h11 =
V1
= 2V
I1 V =0
Fig. 16.48
2
h21 =
I2
I1
; I 2 = −I1
V2 =0
∴ h21 5 – 1
When
∴
I1 = 0; h12 =
V1
V2
; h22 =
I1 =0
V1 5 I2 4, V2 5 I2 4
1
h12 = 1 h22 = 
4
I2
V2
I1 =0
Fig. 16.49
V2
762
Circuits and Networks
PROBLEM 16.6
For the hybrid equivalent circuit shown in Fig. 16.50, (a) determine the current gain, and (b) determine the
voltage gain.
Fig. 16.50
Solution From the port 2-29 we can find
I2 =
(25I1 )(0.05×106 )
(1500 + 0.05×106 )
I2
1.25×106
=
= 24.3
I1 0.0515×106
(a) Current gain
(b) Applying KVL at the port 1-19,
V1 5 500 I1 1 2 3 10–4 V2
V1 − 2×10−4 V2
500
Applying KCL at the port 2-29,
V
I 2 = 25 I1 + 2 ×10−6
0.05
I1 =
also
I2 =
−V2
1500
−V2
V
= 25 I1 + 2 ×10−6
1500
0.05
Substituting the value of I1 from Eq. (16.33), in the above equation, we get
V − 2×10−4 V  V
−V2
2
2
= 25  1
×10−6
 +

1500
500
0.05
– 6.6 3 10–4 V2 5 0.05V1 – 0.1 3 10–4 V2 1 0.2 3 10–4 V2
∴
V2
= −73.89
V1
The negative sign indicates that there is a 180° phase shift between input and output voltages.
(16.33)
Two-Port Networks
763
PROBLEM 16.7
The hybrid parameters of a two‑port network shown in Fig. 16.51 are h11 5 1 K; h12 5 0.003; h21 5 100;
h22 5 50 m. Find V2 and Z‑parameters of the network.
Fig. 16.51
Solution
V1 5 h11 I1 1 h12 V2
(16.34)
I2 5 h21 I1 1 h22 V2
(16.35)
At the port 2-29, V2 5 – I2 2000
Substituting in Eq. (16.35), we have
I2 5 h21I1 – h22I2 2000
I2 (1 1 h22 2000) 5 h21 I1
I2(1 1 50 3 10–6 3 2000) 5 100 I1
I2 =
100 I1
1.1
Substituting the value of V2 in Eq. (16.34), we have
V1 5 h11 I1 – h12 I2 2000
Also, at the port 1-19,
∴
V1 5 VS – I1 500
Vs − I1 500 = h11 I1 − h12
100 I1
× 2000
1.1
(10×10−3 ) − 500 I1 = 1000 I1 − 0.003×100 I1 × 2000
1.1
954.54I1 5 10 3
10–3
I1 5 10.05 3 10–6 A
V1 5 VS – I1 3 500
5 10 3 10–3 – 10.5 3 10–6 3 500 5 4.75 3 10–3 V
V2 =
V1 − h11 I1
h12
764
Circuits and Networks
V2 =
4.75×10−3 −1000 ×10.5×10−6
= −1.916 V
0.003
(b) Z-parameters of the network can be found from Table 16.1.
Z11 =
∆h
h h − h21h12 1×103 ×50 ×10−6 −100 × 0.003
= 11 22
=
h22
h22
50 ×10−6
5 – 5000 V
Z12 =
h12
0.003
=
= 60 V
h22 50 ×10−6
Z 21 =
−h21
−100
=
= −2×106 V
h22
50 ×10−6
Z 22 =
1
= 20 ×103 V
h22
PROBLEM 16.8
The Z‑parameters of a two‑port network shown in Fig. 16.52 are Z11 5 Z22 5 10 V; Z21 5 Z12 5 4 V. If the
source voltage is 20 V, determine I1, V2, I2, and input impedance.
Fig. 16.52
Solution Given V1 5 VS 5 20 V
From Section 16.11.1,

Z Z 
V1 = I1  Z11 − 12 21 

Z L + Z 22 
where ZL 5 20 V
∴

4× 4 
20 = I1 10 −


20 + 10 
I1 5 2.11 A
I 2 = −I1
Z 21
4
= −2.11×
= −0.281
Z L + Z 22
20 + 10
765
Two-Port Networks
At the port 2-29,
V2 5 – I2 3 20 5 0.281 3 20 5 5.626 V
Input impedance 5
V1
20
=
= 9.478 V
I1 2.11
PROBLEM 16.9
The Y‑parameters of the two‑port network shown in Fig. 16.53 are Y11 5 Y22 5 6  ; Y12 5 Y21 5 4 
Determine the driving‑point admittance at the port 2‑29 if the source voltage is 100 V and has an impedance
of 1 ohm.
Fig. 16.53
Solution From Section 16.11.2,
I 2 Y22Ys + Y22Y11 − Y21Y12
=
V2
Ys + Y11
where YS is the source admittance 5 
6 ×1 + 6 ×6 − 4 × 4
= 3.714 
1+ 6
1
Or the driving-point impedance at the port 2-29 5
V
3.714
∴ the driving-point admittance =
PROBLEM 16.10
Obtain the Z‑parameters for the two‑port unsymmetrical
lattice network shown in Fig. 16.54.
Solution From Section 16.12, we have
Z11 =
( Z a + Z b )( Z d + Z c )
Z a + Zb + Zc + Z d
=
(1 + 3)(2 + 5)
= 2.545 V
1+ 3 + 5 + 2
Zb Zc − Z a Z d
3×5 −1× 2
= 1.181 V
=
Z a + Zb + Zc + Z d
11
Z21 5 Z12
Z 21 =
Z 22 =
( Z a + Z c )( Z d + Z b )
Z a + Zb + Zc + Z d
=
(1 + 5)(2 + 3)
= 2.727 V
11
Fig. 16.54
766
Circuits and Networks
PROBLEM 16.11
For the ladder two‑port network shown in Fig. 16.55, find the open‑circuit driving‑point impedance at the
port 1‑2.
Fig. 16.55
Solution The Laplace transform of the given network is shown in Fig. 16.56.
Fig. 16.56
Then the open-circuit driving-point impedance at the port 1-2 is given by
1
Z11 = ( s + 1) +
1
s+
1
( s + 1) +
s+
1
( s + 1) +
=
1
s
s 6 + 3s 5 + 8s 4 + 11s 3 + 11s 2 + 6 s + 1
s 5 + 2 s 4 + 5s 3 + 4 s 2 + 3s
PROBLEM 16.12
For the bridged T‑network shown in Fig. 16.57, find the driving‑ point admittance y11 and transfer admittance
y21 with a 2 V load resistor connected across the port 2.
Two-Port Networks
Fig. 16.57
Solution The corresponding Laplace transform network is shown in Fig. 16.58.
Fig. 16.58
The loop equations are
 1
1
I1 1 +  + I 2   − I 3 = V1
 s
 s
1
 1
I1   + I 2 1 +  + I 3 = 0
 s
 s

1
I1 (−1) + I 2 + I 3 2 +  = 0

s
Therefore,
 1 
1 + 
 s 
D=
1
s
−1
Similarly,
D11 =
1
s
1+
−1
1
s
1
 1 
1 + 
 s 
1
s
1


2 + 1 

s 
=
1
2+
=
s+2
s2
1
s
s 2 + 3s + 1
s2
767
768
Circuits and Networks
1
s
+1
=
s 2 + 2s +1
and
D12 =
Hence,
Y11 =
D11 s 2 + 3s + 1
=
s+2
D
and
Y21 =
D12 −( s 2 + 2 s + 1)
=
D
s+2

1
+1 2 + 

s
s2
PROBLEM 16.13
For the two‑port network shown in Fig. 16.59, determine the h‑parameters. Using these parameters, calculate
the output (Port 2) voltage, V2, when the output port is terminated in a 3 V resistance and a 1V(dc) is applied
at the input port (V1 5 1 V).
Fig. 16.59
Solution The h-parameters are defined as
V1   h11
 =
 I 2   h21
  
h12   I1 
 
h22  V2 
For V2 5 0, the circuit is redrawn as shown in Fig. 16.60
(a).
h11 =
i ×1 + 3i1
V1
= 1
=4
i1
I1 V =0
2
h21 =
Fig. 16.60 (a)
I2
I1
=
V2 =0
For I1 5 0, the circuit is redrawn as shown in Fig. 16.60 (b).
h12 =
V1
I
1
= 1; h22 = 2 = = 0.5
V2
V2 2
i2 2i1 − i1
=
=1
i1
i1
769
Two-Port Networks
Hence,
4 1 

h=
 1 0.5


V1 5 1 V
V1 5 4I1 1 V2
I2 5 I1 1 0.5 V2
Eliminating I1 from the above equations and putting
−V
−3
V1 = 1 and I 2 = 2 , we get V2 =
V
3
7
Fig. 16.60 (b)
PROBLEM 16.14
Find the current transfer ratio I 2 for the network shown in Fig. 16.61.
I1
Fig. 16.61
Solution By transforming the current source into voltage source, the given circuit can be redrawn as shown
in Fig. 16.62.
Applying Kirchhoff’s nodal analysis,
V1 − ( I1 + 2 I 3 )
V1 V1 −V2
+
=0
1
2
V2 −V1 I1
and
− − I2 = 0
2
2
Putting V1 5 – I3 and V2 5 – I2
The above equations becomes
I − I3
−I 3 − I1 − 2 I 3 − I 3 + 2
=0
2
I 2 − I 3 I1
and
− − I2 = 0
2
2
or I1 0.5I2 – 4.5 I3 5 0
and – 0.5 I1 – 1.5I2 1 0.5I3 5 0
By eliminating I3, we get
1
+
I 2 −5.5
=
= −0.42
I1
13
Fig. 16.62
770
Circuits and Networks
PROBLEM 16.15
Obtain Y‑parameters of the two‑port network shown in Fig. 16.63.
Fig. 16.63
Solution The above network is the parallel connection of two-port networks. Y-parameters of such networks
can be found by finding individual Y-parameters of the respective networks.
T-and p-networks of the above figure are shown separately.
(a)
(b)
Fig. 16.64
Y-parameters of the T-network are given by
6
5
−4
Y11 = ; Y22 = ; Y21 = Y12 =
7
7
7
The Y-parameters of the p-network are given by
Y11 5 2; Y12 5 1; Y22 5 3; Y21 5 21
Y-parameters of the combination are given by
6
20
5
26
Y11 = + 2 = , Y22 = + 3 =
7
7
7
7
−4
−5
Y12 = Y21 =
−1 =
7
7
PROBLEM 16.16
Find the transmission parameters for the network shown in Fig. 16.65.
Fig. 16.65
Two-Port Networks
Solution Equations of transmission parameters are
V1 = AV2 − BI 2
I1 = CV2 − DI 2
A=
C=
V1
V2
I1
V2
; −B =
I 2 =0
V1
I2
V2 =0
I1
I2
V2 =0
; −D =
I 2 =0
When I2 5 0; V1 5 I1 1 4I1 5 5I1
and 2V223I1 1 4I1 5 0
V2 = I1 ⇒
I1
=1
I2
∴ V1 = 5V2 ⇒
V1
=5
V2
When V2 5 0; the network is shown in Fig. 16.66.
Fig. 16.66
Loop equations V1 5 5I1 1 4I2
23I1 1 2I2 1 4I2 1 4I1 5 0
from which, I1 = −6 I 2 ⇒
I1
= −6
I2
∴ V1 5 230I2 1 4I2
5 226I2
V1
= −26
I2
A=
−V
V1
= 5; B = 1 = 26
V2
I2
C=
−I
I1
= 1; D = 1 = 6
V2
I2
771
772
Circuits and Networks
PROBLEM 16.17
Determine Z‑ and Y‑parameters for the circuit shown in Fig. 16.67.
Fig. 16.67
Solution Z‑parameters
Let I3 be the current in 2 V resistor
V1 = 2 I 3 ⇒ I 3 = V1 / 2
Applying KVL to the outer loop,
2 I 3 − ( I1 − I 3 ) − 3V1 − ( I1 + I 2 − I 3 ) = 0
3V1 = −2 I1 − I 2 + 4 I 3
V2 = I1 − I 3 + I 2
V2 = I1 + I 2 −
3V1 = −2 I1 − I 2 + 2V1
V1 = −2 I1 − I 2
1
V2 = I1 + I 2 − (−2 I1 − I 2 )
2
3
V2 = 2 I1 + I 2
2
3
∴ Z11 = −2; Z12 = −1; Z 21 = 2; Z 22 =
2
Y‑parameters
From the above equations,
V2 + V1 = I1 + I 2 −
V1
− 2 I1 − I 2
2
−3
V1 −V2
2
Multiply equations V2 with 2 and add the equation V1
I1 =
2V2 + V1 = 2 I1 + 2 I 2 −V1
2V2 + 2V1 = 2 I1 + 2 I 2
Also V1 = −2 I1 − I 2
V1
2
773
Two-Port Networks
2V2 + 3V1 = I 2
∴
Y11 =
−3

2
Y12 = −1 
Y21 = 3 
Y22 = 2 
PROBLEM 16.18
Find Z21 and Z22 for the network shown in Fig. 16.68.
+
I2
I1
+
Fig. 16.68
Solution Transforming the dependent current source into voltage source, the network is shown as follows:
−
−
−
Fig. 16.69
Let I3 be the current through 2 V. KVL to the outer loop,
2V2 1 2V1 1 I2 2 I3 + V15 0
2V2 1 3V1 1 I2 2 I3 5 0
Also, 2V1 1 (I1 1 I2 2 I3) 1 2V2 5 0
V1 5 I1 1 I2 2 I3 1 2V2
from which
27V2 2 3I1 2 2I2 1 2I3 5 0
V
where I 3 = 2
2
−I1 I 2
∴ V2 =
−
2
3
−1
−1
; Z 22 =
Hence, Z 21 =
2
3
774
Circuits and Networks
PROBLEM 16.19
Obtain the transmission parameters for the following T‑network and verify the reciprocity theorem.
Fig. 16.70
Solution
V1 5 AV2 2 BI2
I1 5 CV2 2 DI2
Fig. 16.71
When I2 5 0,
V1 5 I1 (8 1 2j)
V2 5 I1 (324j)
A=
V1
I (8 + 2 j ) 8 + 2 j
= 1
=
= 0.64 + 1.52 j
V2
I1 (3 − 4 j ) 3 − 4 j
C=
I1
I1
1
=
=
= 0.12 + 0.16 j 
V2
I1 (3 − 4 j ) 3 − 4 j
When V2 5 0,
6j
V1
−4j
Fig. 16.72
Two-Port Networks
B=
−I 2 =
−V1
I2
I1 (3 − 4 j )
6−4 j
2I2 5 I1 (0.65 2 0.23j)
V1 5 I1{(5 1 6j) 1 [(3 2 4j)||3]}
= I1[6.96 1 5.3j]
B=
−V1
I (6.96 + 5.3 j )
= 1
I2
I1 (0.65 − 0.23 j )
∴ B 5 6.95 1 10.61j
D = −I1 / I 2 =
I1
= 1.367 + 0.48 j
I1 (0.65 − 0.23 j )
Reciprocity condition is satisfied when AD 2 BC 5 1.
(0.64 + 1.52j ) (1.367+0.48j ) − (6.95+10.61j )(0.12+0.16j )
 


= 1.00 −1.6 ×10−4 = 1.008 −0.009  1
Hence, the condition of reciprocity is verified.
PSpice Problems
PROBLEM 16.1
Using PSpice, find the Z‑parameters for the circuit in Fig. 16.73.
Fig. 16.73
V1 5 Z11 I1 1 Z12 I2
V2 5 Z21 I1 1 Z22 I2
* DETERMINATION OF Z PARAMETERS
.SUBCKT AMP 1 3
R1
1
6
1
775
776
Circuits and Networks
R2
6
3
2
R3
0
6
2
R4
0
3
4
.ENDS
I1
0
1
1
I2
0
3
0
I3
0
5
1
I4
0
4
0
X1
1
3 AMP
X2
4
5 AMP
.OP
.END
**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE 5 27.000 DEG C
****************************************************************
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
(1) 2.5000 (3) 1.0000
(4) 1.0000
(5) 2.0000
(X1.6) 1.5000
(X2.6) 1.0000
Result
Subcircuit 1 is with I2 5 0 (I2 5 0)
output port is open-circuited.
V
Z11 = 1 = 2.5 Ω
I1
Z 21 =
V3
=1Ω
I1
Subcircuit 2 is with I4 5 0
input port open-circuited.
Z12 =
V4 V4
=
=1 V
1
I3
Z 22 =
V5
=2V
I3
PROBLEM 16.2
Using PSpice, find the transmission parameters for the circuit shown in Fig. 16.74.
Fig. 16.74
Two-Port Networks
V1 5 AV2 – BI2
V1 5 CV2 – DI2
* TRANSMISSION PARAMETERS
.SUBCKT AMP 1 2
R1
1
2
2
R2
1
0
2
R3
2
0
2
.ENDS
I1A
0
1
1
I2A
0
2
0
I1B
0
3
1
V2B
4
0
0
X1
1
2 AMP
X2
3
4 AMP
.OP
.END
**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE 5 27.000 DEG C
****************************************************************
NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
(1) 1.3333
(2) .6667
(3) 1.0000
(4) 0.0000
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V2B
5.000E – 01
Result
Subcircuit 1 is with output node open-circuited.
A=
V1 1.333
=
=2
V2 0.667
C=
I1
1
=
= 1.5 
V2 0.667
Subcircuit 2 is with output node short-circuited
B=
−V3 −1
=
=2V
0.5
I2
D=
−I1 −1
=
=2
I2
0.5
777
778
Circuits and Networks
Answers to Practice Problems
16-2.1
Z11 =
YB + YC
Y + YC
Y
; Z12 = Z 21 = C ; Z 22 = A
DY
DY
DY
DY 5 YA YB 1 YB YC 1 YCYA
16-2.5
 5.71 −4.29


 2.14 2.14 


16-2.6
A9 5 3; B9 5 2; C9 5 4; D9 5 3
16-2.7 Y11 5 (0.5 – j0.2)10–3;
Y12 5 Y21 5 ( j0.2 3 10–3)
Y22 5 j(0.02 3 10–3)
16-2.11
Y115 −0.5
Y12 = 0.25
Y22= 0.2
16-2.12
 0.75 0.25
 1.5 −0.5
; Y = 

Z=
 0.25 0.75
−0.5
1.5 



16-2.13
11
16-4.1
2
= ;
3
12
=
5
−1
−1
; Y21 = ; Y22 =
3
3
3
4
1
2
−2
h11 = ; h21 =
; h22 = ; h12 =
3
3
6
3
1
g11 = ; g12 = −1; g21 = 1; g22 = 2
4
16-4.2
Zi 5 1.5 kV; Z0 5 0.033 3 10–3 V
16-4.3
 0.857∠− 31°kV
0.17∠59° 


 8.58∠− 32.1° 1.89∠61.1°m


16-7.1
2 R2 − 0.8 R3
−R2
; Y2 =
DZ
DZ
R1 + R2
R2
R1
Y3 =
; DZ =
R2 − 0.2 R3 R2 + R3
DZ
Y1 =
779
Two-Port Networks
Objective-Type Questions
rrr 16.1
A two-port network is simply a network inside a black box, and the network has only
(a) two terminals
(b) two pairs of accessible terminals
(c) two pairs of ports
rrr 16.2
The number of possible combinations generated by four variables taken two at a time in a two-port network
is
(a) four
(b) two
(c) six
rrr 16.3
What is the driving-point impedance at port one with port
two open- circuited for the network in Fig. 16.75?
(a) 4 V
(b) 5 V
rrr 16.4
What is the transfer impedance of the two-port network
shown in Practice Problem 1 of LO 10?
(a) 1 V
(c) 3 V
rrr 16.5
rrr 16.9
(b) Z12 5 Z21
(c) Z11 5 Z12
(c) –4 
If the two-port network in Practice Problem 5 of LO 2 is
reciprocal then
(a) Y11 5 Y22
(c) Y12 5 Y11
rrr 16.8
Fig. 16.75
What is the transfer admittance of the network shown in
Fig. 16.76.
(a) –2 
(b) –3 
rrr 16.7
(b) 2 V
If the two-port network in Practice Problem 1 of LO 10 is reciprocal or bilateral then
(a) Z11 5 Z22
rrr 16.6
(c) 3 V
Fig. 16.76
(b) Y12 5 Y22
In describing the transmission parameters,
(a) the input voltage and current are expressed in terms of output voltage and current
(b) the input voltage and output voltage are expressed in terms of output current and input current
(c) the input voltage and output current are expressed in terms of input current and output voltage
If Z11 5 2 V; Z12 5 1 V; Z21 5 1 V; and Z22 5 3 V, what is the determinant of admittance matrix?
(a) 5
(b) 1/5
(c) 1
rrr 16.10 For a two-port bilateral network, the three transmission parameters are given by
, what is the value of D?
(a) 1
(b) 1
5
(c)
7
5
6
17
1
A = ; B = and C =
5
5
5
780
Circuits and Networks
rrr 16.11 The impedance matrices of two two-port networks are given by
 3 2
15 5 

 and 
 . If the two networks
 2 3
 5 25




are connected in series, what is the impedance matrix of the combination?
18 7 


 7 28
3
5


 2 25
(b) 
(a) 
15 2


 5 3
(c) 
rrr 16.12 The admittance matrices of two two-port networks are given by
 1 / 2 −1 / 4
 1
−1 / 2 

 and 
 . If
−1 / 4 5 / 8 
−1 / 2 5 / 4 




the two networks are connected in parallel, what is the admittance matrix of the combination?
 1
−1 / 2 

−1 / 2 5 / 4 


(a) 
 2 −1 

−1 5 / 2


(b) 
 3 / 2 −3 / 4 

−3 / 4 15 / 8 


(c) 
rrr 16.13 If the Z-parameters of a two-port network are Z11 5 5 V Z22 5 7 V; Z12 5 Z21 5 3 V then the A, B, C, D
parameters are respectively given by
(a)
5 26 1 7
; ; ;
3 3 3 3
(b)
10 52 2 14
; ; ;
3 3 3 3
(c) 15 ; 78 ; 3 ; 21
3
3 3 3
rrr 16.14 For a symmetric lattice network, the value of the series impedance is 3 V and that of the diagonal impedance
is 5 V, then the Z-parameters of the network are given by
(a) Z11 5 Z22 5 2 V
Z12 5 Z21 5 1/2 V
(b) Z11 5 Z22 5 4 V
Z12 5 Z21 5 1 V
(c) Z11 5 Z22 5 8 V
Z12 5 Z21 5 2 V
rrr 16.15 For a two-port network to be reciprocal,
(a) Z11 5 Z22
(b) y21 5 y22
(c) h21 5 – h12
(d) AD – BC 5 0
rrr 16.16 Two-port networks are connected in cascade. The combination is to be represented as a single two-port
network. The parameters of the network are obtained by adding the individual
(c) A1 B1 C1 D1 matrix
(d) ABCD-parameter matrix
(a) Z-parameter matrix
(b) h-parameter matrix
rrr 16.17 The h-parameters h11 and h12 are obtained
(a) by shorting output terminals
(b) by opening input terminals
(c) by shorting input terminals
(d) by opening output terminals
rrr 16.18 Which parameters are widely used in transmission-line theory?
(a) Z-parameters
(b) Y-parameters
(c) ABCD-parameters
For interactive quiz with answers,
scan the QR code given here
OR
visit
http://qrcode.flipick.com/index.php/274
(d) h-parameters
17
LEARNING OBJECTIVES
17.1
CLASSIFICATION OF FILTERS
Wave filters were first invented by G A Campbell and O I Lobel of the Bell Telephone
LO 1
Laboratories. A filter is a reactive network that freely passes the desired bands of
frequencies while almost totally suppressing all other bands. A filter is constructed
from purely reactive elements, for otherwise the attenuation would never become
zero in the passband of the filter network. Filters differ from simple resonant circuits in providing a substantially
constant transmission over the band which they accept; this band may lie between any limits depending on the
design. Ideally, filters should produce no attenuation in the desired band, called the transmission band or pass
band, and should provide total or infinite attenuation at all other frequencies, called attenuation band or stop
band. The frequency which separates the transmission band and the attenuation band is defined as the cut-off
frequency of the wave filters, and is designated by fc.
Filter networks are widely used in communication systems to separate various voice channels in carrier
frequency telephone circuits. Filters also find applications in instrumentation, telemetering equipment, etc.
where it is necessary to transmit or attenuate a limited range of frequencies.
A filter may, in principle, have any number of pass bands separated by attenuation bands. However, they
are classified into four common types, viz. low-pass, high-pass, band-pass and band-elimination.
17.1.1 Decibel and Neper
The attenuation of a wave filter can be expressed in decibels or nepers. Neper is defined as the natural
logarithm of the ratio of input voltage (or current) to the output voltage (or current), provided that the network
is properly terminated in its characteristic impedance Z0.
782
Circuits and Networks
From Fig. 17.1 (a), the number of
nepers, N = loge
Two port
Network
V1
I
or loge 1 .
V2
I2
A neper can also be expressed in terms of input power, P1
and the output power P2 as N 5 1/2 loge P1/P2.
A decibel is defined as ten times the common logarithms of
the ratio of the input power to the output power.
Fig. 17.1 (a)
∴ Decibel D = 10 log10
P1
P2
The decibel can be expressed in terms of the ratio of input voltage (or current) and the output voltage (or
current.)
D = 20 log10
V1
I
= 20 log10 1
V2
I2
∴ One decibel is equal to 0.115 N.
17.1.2 Low-pass Filter
By definition, a low-pass (LP) filter is one which passes without attenuation all frequencies up to the cut-off
frequency fc, and attenuates all other frequencies greater than fc. The attenuation characteristic of an ideal LP
filter is shown in Fig. 17.1 (b). This transmits currents of all frequencies from zero up to the cut-off frequency.
The band is called pass band or transmission band. Thus, the pass band for the LP filter is the frequency range
0 to fc. The frequency range over which transmission does not take place is called the stop band or attenuation
band. The stop band for an LP filter is the frequency range above fc.
Pass
band
band
Low-Pass filter
band
band
Pass
band
band
band
High-Pass filter
band
band band
band
Band-Pass filter
-
Fig. 17.1 (b)
filter
783
Filters and Attenuators
17.1.3 High-Pass Filter
A high-pass (HP) filter attenuates all frequencies below a designated cut-off frequency, fc, and passes all
frequencies above fc. Thus, the pass band of this filter is the frequency range above fc, and the stop band is the
frequency range below fc. The attenuation characteristic of an HP filter is shown in Fig. 17.1 (b).
17.1.4 Band-Pass Filter
A band-pass filter passes frequencies between two designated cut-off frequencies and attenuates all other
frequencies. It is abbreviated as BP filter. As shown in Fig. 17.1 (b), a BP filter has two cut-off frequencies
and will have the pass band f2 – f1; f1 is called the lower cut-off frequency, while f2 is called the upper cut-off
frequency.
17.1.5 Band-Elimination Filter
A band-elimination filter passes all frequencies lying outside a certain range, while it attenuates all frequencies
between the two designated frequencies. It is also referred as band stop filter. The characteristic of an ideal
band elimination filter is shown in Fig. 17.1 (b).
All frequencies between f1 and f2 will be attenuated while frequencies below f1 and above f2 will be passed.
Frequently Asked Questions linked to LO 1*
rrr 17-1.1
What are the properties of filters?
What are the classifications of filter? Discuss them briefly.
rrr 17-1.3 Explain the concept of insertion loss.
rrr 17-1.2
17.2
[JNTU Nov. 2012]
[JNTU Nov. 2012]
[PU 2012]
FILTER NETWORKS
Ideally a filter should have zero attenuation in the pass band. This condition can
only be satisfied if the elements of the filter are dissipationless, which cannot be
realised in practice. Filters are designed with an assumption that the elements of
the filters are purely reactive. Filters are made of symmetrical T, or p sections. T
and p sections can be considered as combinations of unsymmetrical L sections as
shown in Fig. 17.2.
LO 2
*For answers to Frequently Asked Questions, please visit the link http://highered.mheducation.com/sites/9339219600
784
Circuits and Networks
Fig. 17.2
The ladder structure is one of the commonest forms of filter network. A cascade connection of several T
and p sections constitutes a ladder network. A common form of the ladder network is shown in Fig. 17.3.
Figure 17.3 (a) represents a T-section ladder network, whereas Fig. 17.3 (b) represents the p-section
ladder network. It can be observed that both networks are identical except at the ends.
Fig. 17.3
17.3
EQUATIONS OF FILTER NETWORKS
LO 2
The study of the behaviour of any filter requires the calculation of its propagation constant g, attenuation a,
phase shift b, and its characteristic impedance Z0.
17.3.1 T-Network
Consider a symmetrical T-network as shown in Fig. 17.4.
As has already been mentioned in Section 16.13, if the image
impedances at the port 1-19 and port 2-29 are equal to each
other, the image impedance is then called the characteristic, or
the iterative impedance, Z0. Thus, if the network in Fig. 17.4 is
terminated in Z0, its input impedance will also be Z0. The value
of input impedance for the T-network when it is terminated in Z0
Fig. 17.4
785
Filters and Attenuators
is given by
Z

Z 2  1 + Z0 
Z
2

Zin = 1 +
Z1
2
+ Z 2 + Z0
2
Zin = Z0
also
Z

2 Z 2  1 + Z0 
Z1
2

Z0 = +
2 Z1 + 2 Z 2 + 2 Z0
∴
Z0 =
Z0 =
Z1 ( Z1Z 2 + 2 Z 2 Z0 )
+
2 Z1 + 2 Z 2 + 2 Z0
Z12 + 2 Z1Z 2 + 2 Z1Z0 + 2 Z1Z 2 + 4 Z0 Z 2
2( Z1 + 2 Z 2 + 2 Z0 )
4 Z02 = Z12 + 4 Z1Z 2
Z02 =
Z12
+ Z1Z 2
4
The characteristic impedance of a symmetrical T-section is
Z12
(17.1)
+ Z1Z 2
4
Z0T can also be expressed in terms of open-circuit impedance Z0c and short circuit impedance Zsc of the
Z
T-network. From Fig. 17.4, the open-circuit impedance Z0 c = 1 + Z 2 and
2
Z1
× Z2
Z
Z sc = 1 + 2
Z1
2
+ Z2
2
Z 0T =
Z sc =
Z12 + 4 Z1Z 2
2 Z1 + 4 Z 2
Z0 c × Z sc = Z1Z 2 +
= Z02T
Z12
4
or
Z0T = Z0 c Z sc
(17.2)
‰ Propagation‰Constant‰of‰T-Network‰ By definition,
the propagation constant g of the network in Fig. 17.5 is given
by g 5 loge I1/I2
Fig. 17.5
786
Circuits and Networks
Writing the mesh equation for the second mesh, we get

Z
I1Z 2 = I 2  1 + Z 2 + Z0 

2
Z1
+ Z 2 + Z0
I1
= 2
= eg
I2
Z2
Z1
+ Z 2 + Z0 = Z 2 e g
2
Z
Z0 = Z 2 (e g −1) − 1
2
∴
(17.3)
The characteristic impedance of a T-network is given by
Z12
+ Z1Z 2
4
Z 0T =
(17.4)
Squaring Eqs (17.3) and (17.4) and subtracting Eq. (17.4) from Eq. (17.3), we get
Z 22 (e g −1) 2 +
Z12
Z2
− Z1Z 2 (e g −1) − 1 − Z1Z 2 = 0
4
4
Z 22 (e g −1) 2 − Z1Z 2 (1 + e g −1) = 0
Z 22 (e g −1) 2 − Z1Z 2 e g = 0
Z 2 (e g −1) 2 − Z1e g = 0
(e g −1) 2 =
Z1e g
Z2
e 2 g + 1 − 2e g =
Z1
Z 2 e−g
Rearranging the above equation, we have
e−g (e 2g + 1− 2e g ) =
Z1
Z2
(e g + e−g − 2) =
Z1
Z2
Dividing both sides by 2, we have
Z
e g + e−g
= 1+ 1
2
2Z2
cosh g = 1 +
Z1
2Z2
(17.5)
Still another expression may be obtained for the complex propagation constant in terms of the hyperbolic
tangent rather than hyperbolic cosine.
787
Filters and Attenuators
sinh g = cos h2 g −1
2
2

Z1 
Z1  Z1 

= 1 +
+
 −1 =

 2Z 
Z1  2Z 2 
2
sinh g =
1
Z2
Z1Z 2 +
Z
Z12
= 0T
4
Z2
(17.6)
Dividing Eq. (17.6) by Eq. (17.5), we get
Z 0T
tanh g =
Z2 +
Z2 +
But
Z1
2
Z1
= Z 0c
2
Also, from Eq. (17.2), Z 0T = Z 0c Z sc
tanh g =
Z sc
Z 0c
1
g
=
(cosh g −1)
2
2
Also,
sinh
where
cosh g = 1 + ( Z1 / 2 Z2 )
=
Z1
4Z 2
(17.7)
17.3.2 p-Network
Consider an asymmetrical p-section shown in Fig. 17.6. When the network is terminated in Z0 at the port
2-29, its input impedance is given by

2Z 2 Z0 

2 Z 2  Z1 +

2 Z 2 + Z 0 
Z in =
2Z 2 Z 0
Z1 +
+ 2Z 2
2Z 2 + Z 0
By definition of characteristic impedance, Zin 5 Z0

2Z 2 Z0 

2 Z 2  Z1 +

2 Z 2 + Z 0 
Z0 =
2Z 2 Z 0
Z1 +
+ 2Z 2
2Z 2 + Z 0
Fig. 17.6
788
Circuits and Networks
2 Z 2 Z02
2 Z ( 2 Z1Z 2 + Z0 Z1 + 2 Z0 Z 2 )
+ 2Z0 Z 2 = 2
2Z 2 + Z0
( 2Z 2 + Z0 )
Z0 Z1 +
2 Z0 Z1Z 2 + Z1Z02 + 2 Z02 Z 2 + 4 Z 22 Z0 + 2 Z 2 Z02
= 4 Z1Z 22 + 2 Z0 Z1Z 2 + 4 Z0 Z 22
Z1Z02 + 4 Z 2 Z02 = 4 Z1Z 22
Z02 ( Z1 + 4 Z 2 ) = 4 Z1Z 22
Z02 =
4 Z1Z 22
Z1 + 4 Z 2
Rearranging the above equation leads to
Z0 =
Z1Z 2
1 + Z1 / 4 Z 2
(17.8)
which is the characteristic impedance of a symmetrical p-network,
Z1Z 2
Z 0p =
From Eq. (17.1),
Z1Z 2 + Z12 / 4
Z 0T =
∴ Z0 p =
Z12
+ Z1Z2
4
Z1Z2
Z 0T
(17.9)
Z0p can be expressed in terms of the open-circuit impedance Z0c and short-circuit impedance Zsc of the p
network shown in Fig. 17.6 exclusive of the load Z0.
From Fig. 17.6, the input impedance at the port 1-19 when the port 2-29 is open is given by
Z 0C =
2 Z 2 ( Z1 + 2 Z 2 )
Z1 + 4 Z 2
Similarly, the input impedance at the port 1-19 when the port 2-29 is short-circuited is given by
Z sc =
2 Z1Z 2
2 Z 2 + Z1
Hence,
Z 0c × Z sc =
4 Z1Z 22
Z1Z 2
=
Z1 + 4 Z 2 1 + Z1 / 4 Z 2
Thus, from Eq. (17.8),
Z0 p = Z0 c Z sc
(17.10)
789
Filters and Attenuators
‰ Propagation‰Constant‰of‰p-Network The propagation constant of a symmetrical p-section is the
same as that for a symmetrical T-section,
i.e.,
cosh g = 1 +
Z1
2Z 2
Frequently Asked Questions linked to LO 2
rrr 17-2.1
Compare a first-order low-pass filter to a second-order low-pass filter in terms of (a) voltage gain
and (b) cut-off frequency, etc.
[BPUT 2008]
rrr 17-2.2
Find the characteristic impedance of a T-section as shown in Fig. Q.2. Verify the value of
impedance with the help of open-and short-circuit impedances.
[PU 2010]
rrr 17-2.3
A symmetrical T-network consisting of pure resistances has open-and short-circuit impedances
Zoc = 800 –0º , and Zsc = 600 –0º . Design a symmetrical T-network.
[PU 2010]
rrr 17-2.4
For a symmetrical T-network, prove that:
[PU 2010]
 N — 1
R1/2 = R0 

 N + 1
 2N 

2
 N — 1
R2 = R0 
rrr 17-2.5
Design a suitable matching half-section to match a
symmetrical T-network with ZOT = 500
to a generator
having internal resistance equal to 200 .
[PU 2012]
rrr 17-2.6
Design a -type attenuator to give 20 dB attenuation and to
have a characteristic impedance of 100 . [PTU 2011-12]
rrr 17-2.7
Calculate image impedance and iterative impedance of the
T-network shown in Fig. Q.7
[PU 2010]
17.4
Fig. Q.2
Fig. Q.7
CLASSIFICATION OF PASS BAND AND STOP BAND
It is possible to verify the characteristics of filters from the propagation
constant of the network. The propagation constant g, being a function of
frequency, the pass band, stop band, and the cut-off point, i.e. the point
of separation between the two bands, can be identified. For symmetrical
T or p-sections, the expression for propagation constant g in terms of the
LO 3
hyperbolic functions is given by Eqs (17.5) and (17.7) in Section 17.3. From Eq. (17.7), sin h
*Note: rrr - Level 1 and Level 2 Category
rrr - Level 3 and Level 4 Category
rrr- Level 5 and Level 6 Category
Z1
g
=
.
2
4 Z2
790
Circuits and Networks
If Z1 and Z2 are both pure imaginary values, their ratio, and hence Z1/4Z2, will be a pure real number. Since
Z1 and Z2 may be anywhere in the range from – j t0 1 j, Z1/4Z2 may also have any real value between the
g
infinite limits. Then sinh = Z1 / 4 Z 2 will also have infinite limits, but may be either real or imaginary
2
depending upon whether Z1/4Z2 is positive or negative.
We know that the propagation constant is a complex function g 5 a 1 jb, the real part of the complex
propagation constant a, is a measure of the change in magnitude of the current or voltage in the network,
known as the attenuation constant. b is a measure of the difference in phase between the input and output
currents or voltages, known as phase shift constant. Therefore, a and b take on different values depending
upon the range of Z1/4Z2. From Eq. (17.7), we have
sinh
 a jb 
a
b
a
b
g
= sinh  +  = sinh cos + j cosh sin


2
2
2
2
2
2
2
=
Z1
4Z 2
‰ Case‰A‰ If Z1 and Z2 are the same type of reactances, then
(17.11)
Z1
is real and equal to say a + x.
4Z 2
The imaginary part of the Eq. (17.11) must be zero.
∴
a
b
sin = 0
2
2
a
b
sinh cos = x
2
2
cosh
(17.12)
(17.13)
a and b must satisfy both the above equations.
Equation (17.12) can be satisfied if b/2 5 0 or np, where n 5 0, 1, 2, ..., then cos b/2 = 1 and sinh a/2
Z1
= x=
4Z 2
That x should be always positive implies that
Z1
Z1
> 0 and a = 2 sinh−1
4 Z2
4 Z2
(17.14)
Since a ≠ 0, it indicates that the attenuation exists.
‰ Case‰B‰ Consider the case of Z1 and Z2 being opposite type of reactances, i.e. Z1/4Z2 is negative, making
Z1 / 4 Z 2 imaginary and equal to say Jx
∴ the real part of Eq. (17.11) must be zero.
sinh
a
b
cos = 0
2
2
(17.15)
cosh
a
b
sin = x
2
2
(17.16)
Filters and Attenuators
791
Both the above equations must be satisfied simultaneously by a and b. Equation (17.15) may be satisfied
when a 5 0, or when b 5 p. These conditions are considered separately hereunder.
1. When a 5 0; from Eq. (17.15), sinh a/2 5 0. And from Eq. (17.16) sin b / 2 = x = Z1 / 4 Z2 . But
the sine can have a maximum value of 1. Therefore, the above solution is valid only for negative Z1/4Z2, and
having maximum value of unity. It indicates the condition of pass band with zero attenuation and follows the
condition as
Z
−1 ≤ 1 ≤ 0
4Z 2
b = 2 sin−1
Z1
4Z 2
(17.17)
2. When b 5 p, from Eq. (17.15), cos b/2 5 0. And from Eq. (17.16), sin b/2 5  1; cosh a/2 =
x = Z1 / 4 Z 2 .
Since cosh a/2 $ 1, this solution is valid for negative Z1/4Z2, and having magnitude greater than, or equal
to unity. It indicates the condition of stop band since a ≠ 0.
−a ≤
Z1
≤ −1
4 Z2
a = 2 cosh−1
Z1
4Z 2
(17.18a)
It can be observed that there are three limits for case A and B. Knowing the values of Z1 and Z2, it is
possible to determine the case to be applied to the filter. Z1 and Z2 are made of different types of reactances,
or combinations of reactances, so that, as the frequency changes, a filter may pass from one case to another.
Case A and (ii) in case B are attenuation bands, whereas (i) in Case B is the transmission band.
The frequency which separates the attenuation band from pass band or vice versa is called cut-off
frequency. The cut-off frequency is denoted by fc, and is also termed nominal frequency. Since Z0 is real in
the pass band and imaginary in an attenuation band, fc is the frequency at which Z0 changes from being real
to being imaginary. These frequencies occur at

Z1

(17.18b)

= 0 or Z1 = 0


4Z 2


Z1

(17.18c)
= −1 or Z1 + 4 Z 2 = 0


4Z 2


The above conditions can be represented graphically, as in Fig. 17.7.
band
band
band
Fig. 17.7
792
7.5
Circuits and Networks
CHARACTERISTIC IMPEDANCE IN THE PASS AND STOP BANDS
LO 3
Referring to the characteristic impedance of a symmetrical T-network, from Eq. (17.1), we have
Z 0T =

Z12
Z 
+ Z1Z 2 = Z1Z 2 1 + 1 
 4 Z 2 
4
If Z1 and Z2 are purely reactive, let Z1 5 jx1 and Z2 5 jx2, then

x 
Z 0T = −x1 x2 1 + 1 
 4 x2 
(17.19)
A pass band exists when x1 and x2 are of opposite reactances and
−1 <
x1
<0
4 x2
Substituting these conditions in Eq. (17.19), we find that Z0T is positive and real. Now consider the stop
band. A stop band exists when x1 and x2 are of the same type of reactances; then x1/4x2 . 0. Substituting these
conditions in Eq. (17.19), we find that Z0T is purely imaginary in this attenuation region. Another stop band
exists when x1 and x2 are of the same type of reactances, but with x1/4x2 , – 1. Then from Eq. (17.19), Z0T is
again purely imaginary in the attenuation region.
Thus, in a pass band if a network is terminated in a pure resistance R0(Z0T 5 R0), the input impedance
is R0 and the network transmits the power received from the source to R0 without any attenuation. In a stop
band, Z0T is reactive. Therefore, if the network is terminated in a pure reactance (Z0 5 pure reactance), the
input impedance is reactive, and cannot receive or transmit power. However, the network transmits voltage
and current with 90° phase difference and with attenuation. It has already been shown that the characteristic
impedance of a symmetrical p-section can be expressed in terms of T. Thus, from Eq. (17.9), Z0p 5 Z1Z2/Z0T.
Since Z1 and Z2 are purely reactive, Z0p is real if Z0T is real, and Z0x is imaginary if Z0T is imaginary. Thus,
the conditions developed for T-sections are valid for p-sections.
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 3*
For a p-section filter network shown in Fig. Q.1. calculate the cut-off frequency and the value
of nominal impedance in the pass band.
rrr 17-3.2 A p-section filter network is shown in Fig. Q.2. Calculate the cut-off frequency and phase shift
at 10 kHz. What is the value of nominal impedance in the pass band?
rrr 17-3.1.
Fig. Q.1
Fig. Q.2
793
Filters and Attenuators
17.6
CONSTANT-K LOW-PASS FILTER
A network, either T or p, is said to be of the constant-k type if Z1 and Z2
of the network satisfy the relation
Z1Z2 5 k 2
LO 4
(17.20)
where Z1 and Z2 are impedances in the T and p-sections as shown in Fig. 17.8. Equation (17.20) states that
Z1 and Z2 are inverse if their product is a constant, independent of frequency. k is a real constant, that is the
resistance. k is often termed as design impedance or nominal impedance of the constant k-filter.
Fig. 17.8
The constant k, T or p-type filter is also known as the prototype because other more complex networks
can be derived from it. A prototype T and p-sections are shown in Fig. 17.8 (a) and (b), where Z1 5 jvL and
Z2 5 1/jvC. Hence, Z1Z 2 =
L
= k 2 which is independent of frequency.
C
Z1Z 2 = k 2 =
L
C
or k =
L
C
(17.21)
Since the product Z1 and Z2 is constant, the filter is a constant-k type. From Eq. (17.18 (a)), the cut-off
frequencies are Z1/4Z2 5 0,
i.e.,
i.e.,
−v2 LC
=0
4
Z1
f = 0 and
= −1
4Z 2
−v2 LC
= −1
4
or
fc =
1
p LC
(17.22)
The pass band can be determined graphically. The reactances of Z1 and 4Z2 will vary with frequency as
drawn in Fig. 17.9. The cut-off frequency at the intersection of the curves Z1 and – 4Z2 is indicated as fc. On
the X-axis, as Z1 5 –4Z2 at the cut-off frequency, the pass band lies between the frequencies at which Z1 5 0,
and Z1 5 – 4Z2. All the frequencies above fc lie in a stop or attenuation band. Thus, the network is called a
low-pass filter.
794
Circuits and Networks
band
band
Fig. 17.9
We also have from Eq. (17.7) that
sinh
From Eq. (17.22),
Z1
J v LC
g
−v2 LC
=
=
=
4
2
2
4 Z2
LC =
1
fc p
f
g
j 2pf
=
= j
fc
2
2pf c
We also know that in the pass band
Z
−1 < 1 < 0
4Z 2
∴
sinh
−v2 LC
<0
4
 f 2
−1 < −  < 0
 f c 
−1 <
or
f
<1
fc
and
 f 
b = 2 sin−1   ; a = 0
 f c 
In the attenuation band,
Z1
f
< −1, i.e. < 1
4Z 2
fc
 Z 
 f 
a = 2 cosh−1  1  = 2 cosh−1   ; b = p
 f c 
 4 Z 2 
The plots of a and b for pass and stop bands are shown
in Fig. 17.10.
Fig. 17.10
795
Filters and Attenuators
Thus, from Fig. 17.10,
 f 
a = 0, b = 2 sinh−1   for f < f c
 f c 
 f 
a = 2 cosh−1  ; b = p for f > f c
 f 
c
The characteristic impedance can be calculated as follows:

Z 
Z0T = Z1Z2 1 + 1 
 4 Z2 
=
L  v2 LC 
1−

C 
4 
 f 2
Z 0T = k 1−  
 f c 
(17.23)
From Eq. (17.23), Z0T is real when f , fc, i.e. in the pass
band at f 5 fc, Z0T 5 0; and for f . fc, Z0T is imaginary in
the attenuation band, rising to infinite reactance at infinite
frequency. The variation of Z0T with frequency is shown in
Fig. 17.11.
Similarly, the characteristic impedance of a p-network
is given by
Z0 p =
Z1Z2
=
Z 0T
k
 f 2
1−  
 f c 
(17.24)
Fig. 17.11
The variation of Z0p with frequency is shown in Fig. 17.11. For f , fc, Z0p is real; at f 5 fc, Z0p is infinite,
and for f . fc, Z0p is imaginary. A low-pass filter can be designed from the specifications of cut-off frequency
and load resistance.
At cut-off frequency, Z1 5 – 4Z2
−4
j vc L =
j vcC
p2fc2LC 5 1
Also, we know that k = L / C is called the design impedance or the load resistance
L
k2 =
∴
C
p2fc2 k2C 2 5 1
C=
1
gives the value of the shunt capacitance
pf c k
and L = k 2 C =
k
gives the value of the series inductance.
pf c
796
Circuits and Networks
EXAMPLE 17.1
Design a low-pass filter (both p and T-sections) having a cut-off frequency of 2 kHz to operate with a
terminated load resistance of 500 V.
L
5 500 V, and fc 5 2000 Hz
C
Solution‰ It is given that k =
We know that
L=
500
k
=
= 79.6 mH
pf c 3.14× 2000
C=
1
1
=
= 0.318 mF
pf c k 3.14 ⋅ 2000 ⋅ 500
The T and p-sections of this filter are shown in Fig. 17.12 (a) and (b) respectively.
Fig. 17.12
17.7
CONSTANT-K HIGH PASS FILTER
LO 4
A constant-K high-pass filter can be obtained by changing the positions of series and shunt-arms of the
networks shown in Fig. 17.8. The prototype high-pass filters are shown in Fig. 17.13, where Z1 5 – j/vC and
Z2 5 jvL.
Fig. 17.13
Again, it can be observed that the product of Z1 and Z2 is independent of frequency, and the filter design
obtained will be of the constant-k type. Thus, Z1Z2 are given by
Z1Z2 =
k=
L
−j
j vL = = k 2
vC
C
L
C
797
Filters and Attenuators
The cut-off frequencies are given by Z1 5 0 and Z1 5 – 4Z2.
j
= 0 , or v → 
Z1 5 0 indicates
vC
From Z1 5 – 4Z2,
−j
= −4 j v L
vC
1
v2 LC =
4
1
fc =
4p LC
or
(17.25)
Fig. 17.14
The reactances of Z1 and Z2 are sketched as functions of
frequency as shown in Fig. (17.14).
As seen from Fig. 17.14, the filter transmits all frequencies between f 5 fc and f 5 . The point fc from the
graph is a point at which Z1 5 – 4Z2.
From Eq. (17.7),
sinh
Z1
g
−1
=
=
2
4Z2
4v2 LC
From Eq. (17.25), f c =
∴
∴
1
4p LC
LC =
sinh
1
4pf c
−( 4p)2 ( f c )2
f
g
=
= j c
2
2
f
4v
In the pass band, −1 <
f 
Z1
f
< 0 , a 5 0 or the region in which c <1 is a pass band b = 2 sin−1  c 
 f 
4Z 2
f
In the attenuation band,
f
Z1
< −1, i.e. c > 1
f
4 Z2
 Z 
a = 2 cosh−1  1 
 4Z 
 2
f 
= 2 cos−1  c  ; b = −p
 f 
The plots of a and b for pass and stop bands of a highpass filter network are shown in Fig. 17.15.
A high-pass filter may be designed similar to the low-pass
filter by choosing a resistive load r equal to the constant k,
such that R = k = L / C
Fig. 17.15
798
Circuits and Networks
fc =
Since
1
4p L / C
fc =
1
k
=
4pL 4pCk
C=
L
,
k
L=
k
1
and C =
4pf c
4pf c k
The characteristic impedance can be calculated using the relation

Z 
L 
1 
Z 0T = Z1Z 2 1 + 1  =

1−
 4 Z 2 
C  4v2 LC 
 f 2
Z 0T = k 1−  c 
 f 
Similarly, the characteristic impedance of a p-network is
given by
Z0 p =
=
Z1Z2
k2
=
Z 0T
Z 0T
k
 f 2
1−  c 
 f 
(17.26)
The plot of characteristic impedances with respect to
frequency is shown in Fig. 17.16.
Fig. 17.16
EXAMPLE 17.2
Design a high-pass filter having a cut-off frequency of 1 kHz with a load resistance of 600 V.
Solution‰ It is given that RL 5 K 5 600 V and fc 5 1000 Hz
∴
L=
600
K
=
= 47.74 mH
4pf c 4 × p×1000
C=
1
1
=
= 0.133 mF
4pkf c 4p×600 ×1000
The T and p-sections of the filter are shown in Fig. 17.17.
799
Filters and Attenuators
Fig. 17.17
NOTE: YOU ARE NOW READY TO ATTEMPT THE FOLLOWING PROBLEMS
Practice Problems linked to LO 4
rrr 17-4.1
Design a high-pass filter with a cut-off frequency of 1 kHz with a terminated design impedance
of 800 V.
Frequently Asked Questions linked to LO 4
rrr 17-4.1
Analyse a prototype low-pass filter with derivation of all necessary equations and also discuss the
different characteristics of the filter.
[BPUT 2007]
rrr 17-4.2 A constant-K low-pass filter is designed to cut-off at a frequency of 1000 Hz and the resistance
of the load circuit is 50 . Calculate the values of the components required and the attenuation
constant per section at a frequency of 1500 Hz.
[BPUT 2008]
rrr 17-4.3 Design a constant-k low-pass filter having fc = 2 kHz and design impedance Ro = 600 ,. Obtain
the value of attenuation at 25 kHz.
[JNTU Nov. 2012]
rrr 17-4.4 Design constant-k low-pass T and -section filter to be terminated in 600
having cut-off
frequency of 3 kHz.
[PTU 2009-10]
rrr 17-4.5 Draw the reactance curve for a constant k low-pass filter and derive the expression for cut-off
frequency and design impedance (R0).
[PU 2012]
rrr 17-4.6 Design a constant-K high-pass filter, having fc = 4 kHz and design impedance Ro = 600 .
[BPUT 2007]
rrr 17-4.7 Can you design a filter (low and high-pass) for a cut-off frequency of 50 Hz. If you can, what is
the value of parameters?
[BPUT 2008]
rrr 17-4.8 A constant-K high-pass filter is required for a cut-off frequency of 1500 Hz. The resistance of the
load circuit is 600 . Determine the values of the components required.
[BPUT 2008]
rrr 17-4.9 Design a T-section constant-K high-pass filter having cut-off frequency of 10 kHZ and design
impedance Ro = 600 ohms. Find its characteristic impedance and constant at 25 kHz.
[JNTU Nov. 2012]
rrr 17-4.10 A prototype high-pass filter has a cut-off frequency of 10 kHz and design impedance of 600 .
Find the values of L and C. Also find attenuation in dB and phase shift in degrees at a frequency
of 8 kHz.
[PU 2010]
17.8
m-DERIVED T-SECTION
It is clear from Figs 17.10 and 17.15 that the attenuation is not sharp in the
stop band for k-type filters. The characteristic impedance, Z0 is a function
of frequency and varies widely in the transmission band. Attenuation can be
LO 5
800
Circuits and Networks
increased in the stop band by using ladder section, i.e. by connecting two or more identical sections. In order
to join the filter sections, it would be necessary that their characteristic impedances be equal to each other at
all frequencies. If their characteristic impedances match at all frequencies, they would also have the same pass
band. However, cascading is not a proper solution from a practical point of view. This is because practical
elements have a certain resistance, which gives rise to attenuation in the pass band also. Therefore, any
attempt to increase attenuation in stop band by cascading also results in an increase of ‘a’ in the pass band. If
the constant-k section is regarded as the prototype, it is possible to design a filter to have rapid attenuation in
the stop band, and the same characteristic impedance as the prototype at all frequencies. Such a filter is called
m-derived filter. Suppose a prototype T-network shown in Fig. 17.18 (a) has the series arm modified as shown
in Fig. 17.18 (b